TEACHING NOTES

PARABOLA
1ST LECTURE
1. CONIC SECTIONS :
A conic section, or conic is the locus of a point which moves in a plane so that the ratio
of its distance from a fixed point to its perpendicular distance from a fixed straight line
PS
is a constant i.e. = constant = e.
PF
The fixed point is called the FOCUS.
The fixed straight line is called the DIRECTRIX.
The constant ratio is called the ECCENTRICITY denoted by e.
The line passing through the focus & perpendicular to the directrix is called the AXIS.
A point of intersection of a conic with its axis is called a VERTEX.

2. GENERAL EQUATION & OF A CONIC : (Focal Directrix Property)

The general equation of a conic with focus ( , ) & directrix lx + my + n = 0 is :
(l2 + m2) [(x  )2 + (y  )2] = e2 (lx + my + n)2  ax2 + 2hxy + by2 + 2gx + 2fy + c = 0
this simplifies to ax2 + 2hxy + by2 + 2gx + 2fy + c = 0.
two cases arise, now
CASE (I) : When The Focus Lies On The Directrix .
In this case D  abc + 2fgh  af2  bg2  ch2 = 0 & the general equation of a conic
represents a pair of straight lines if :
consider the example
e > 1 the lines will be real & distinct intersecting at S.  with y - axis as directrix
e = 1 the lines will coincident passing through S.  and ' O' as focus. Equation
 of locus is y 2  x 2 (e 2  1)
e < 1 the lines will be imaginary.

CASE (II) : When The Focus Does Not Lie On Directrix.

a parabola an ellipse a hyperbola rectangular hyperbola
e=1;D0 , 0<e<1;D0; e>1;D0 ; e>1; D0
h² = ab h² < ab h² > ab h² > ab ; a + b = 0
|
circle
if e = 0 ; a = b and h = 0
Hence the exact nature of conic section depends not only on its eccentricity but also
on the position of focus relative to directrix.
Ex-1 Identify the conic
Ex-2 Locus of a point which moves such that the ratio if its distance from (1, 2) to its
perpendicular distance from 4x – 3y + 2 = 0 is 3 (Locus is a pair of straight lines)

3. STANDARD EQUATION & GENERAL TERMINOLOGY :

Standard equation of parabola is obtained by placing its
focus at (a, 0) and taking its directrix as the line x + a = 0
so that origin lies on the curve PS = PM
(x – a)2 + y2 = (x + a)2
y2 = 4ax (a > 0)
 SOME IMPORTANT DEFINITIONS ARE :
The distance of a point P on the parabola from the focus is called the Focal Distance
Of The Point and is equal to the distance of point P from the directrix. This is also
known as Focal Directrix Property.
A chord of the parabola, which passes through the focus is called a Focal Chord.
A chord of the parabola perpendicular to the axis of the symmetry is called a Double
Ordinate.
A double ordinate passing through the focus or a focal chord perpendicular to
the axis of parabola is called the Latus Rectum. For y² = 4ax.
 Length of the latus rectum = 4a .
 ends of the latus rectum are L(a , 2a) & L’ (a ,  2a).

Note that :
(i) Perpendicular distance from focus to directrix = half the latus rectum.
(ii) Vertex is middle point of the focus & the point of intersection of directrix & axis.
(iii) Point of intersection of axis and directrix is called foot of directrix.
(iv) Two parabolas are laid to be equal if they have the same latus rectum. Four standard
forms of the parabola are, y² = 4ax ; y² =  4ax ; x² = 4ay ; x² =  4ay
Standard parabola y2 = 4ax (x > 0) at a glance

Four Standard Parabolas :

Note : In case the vertex of the parabola is (h, k) then its equation can be taken as
(a) (y – k)2 = 4a(x – h) if its axis is parallel to x-axis.
(b) (x – h)2 = 4a(y – k) if its axis is parallel to y-axis.
4. PARAMETRIC REPRESENTATION : (Most common parametric coordinates)
x = at2 and y = 2at together represents a parabola.

Examples on basic definition and focal directrix property :

Ex-1 Find everything for the parabola
2 2
 5  7
(a) 4y2 + 12x – 20y + 67 = 0 [Ans.  y     3 x   ]
 2  2
(b) x2 + 4y + 2y = 0 [Ans. standard equation : X2 = – 2Y ]
(c) 4x2 + 2y = 8x – 7

Ex-2 Find the equation of the parabola

(a) whose focus is (1, –1) and directrix x + y – 7 = 0.
(b) whose vertex is the point (4, – 3) and whose latus rectum is 4 and whose axis is parallel
to the x-axis. [Ans. y2 – 4x + 6y + 25 = 0]
[Sol. vertex is (4, – 3) and L1L2 = 4
 (y + 3)2 = 4 (x – 4)
y2 + 6y + 9 = 4x – 16
y2 – 4x + 6y + 25 = 0 Ans. ]
(c) passing through the point (– 4, – 7) and whose directrix is parallel to x axis and whose
vertex is the point (4, – 3). [Ans. x2 – 8x + 16y + 64 = 0]
[Sol. Since the directrix is parallel to x-axis hence axis of symmetric is parallel to y-axis
As vertex is (4, – 3) hence the equation of the parabola can be taken as
(x – 4)2 = 4a (y + 3)
since it passes through (– 4, – 7)
6y = – 16a  a = – 4
 (x – 4)2 = – 16(y + 3) Ans. ]
(d) If a variable circle touches a fixed circle and a fixed
line then prove that the locus of the centre of the
variable circle is a parabola whose directrix is parallel
to a given line at a distance equal to the radius of the
given circle.
[Sol. Distance of P from 'O' is equal to PM
= distance of P from the directrix D = R + r
and PO  locus of P is a parabola with 'O' as its focus and 'D' as its directrix]

Ex-3 For some parabola, tangent at the vertex is 3x – 4y = 5 and

its focus is (1, 2). Find everything.

Ex-4 (a) Prove that the area of the triangle whose vertices are (xi , yi ), i = 1, 2, 3 and
1
inscribed in the parabola y2 = 4ax if | (y1 – y2) (y2 – y3) (y3 – y1) |
8a
(b) find the side of an equilateral triangle inscribed in y2 = 4ax if one of its vertex
coincides with the vertex of the parabola. [Ans. 8 3 a ]
1 2
[Hint: tan 30° = 
3 t
 t= 2 3
Hence P is (12a, 4 3 a )  side = 8 3 a ]
Ex-5 A variable  always passes through (1, 0) and touches the curve
y = tan (tan–1x). Find the equation to the locus of its centre.

Ex-6 Obtain the locus of the point of trisection of double ordinates

of the parabola y2 = 4ax.
4a x
[Sol. h = at2 [Ans. y2 = ]
9
6 | k | = 4at  3 | k | = 2at
now eliminate t ]
Ex-7 Prove that the parabolas
y2 = 4b (x – 2a + b) and x2 + 4a (y – 2b – a) = 0 [Q.37, Ex-26, Loney]
intersect orthogonally at the common end of their latus rectum.
[Hint: y2 = 4b (x – (2a – b) ) or y2 = 4bX where x – (2a – b) = X
x2 + 4a (y – (a + 2b) ) or x2 = – 4aY where y – (a + 2b) = Y
for y2 = 4bX, extremities of latus rectum (b, 2b) and (b, – 2b) w.r.t. X Y axis
i.e. (2a, 2b) and (2a, – 2b) w.r.t. xy axis
for x2 = – 4aY, extremities of latus rectum (2a, – a) and (–2a, –a) w.r.t. XY axis
i.e. (2a, 2b) and (–2a, 2b)
Hence the common end of latus rectum (2a, 2b)
dy dy 2b
now for 1st parabola 2y = 4b  = y = 1 at (2a, 2b)
dx dx 1
dy dy x
also for 2nd parabola or2x = – 4a
=– = – 1 at (2a, 2b) ]
dx dx 2a
Ex-8 A variable parabola is drawn to pass through A & B, the ends of a diameter of a given
circle with centre at the origin and radius c & to have as directrix a tangent to a
concentric circle of radius 'a' (a >c) ; the axes being AB & a perpendicular diameter,
prove that the locus of the focus of the parabola is the standard ellipse
x 2 y2
2
 2  1 where b2 = a2 – c2. [Q.22, Ex-25, Loney]
a b
[Sol. (h – c)2 + k2 = (c cos – a)2 ....(1)
2 2
sub. (h + c) + k = (c cos + a) 2 ....(2)
—————————————
4ch = 4ca cos   h = a cos .....(3)
add 2(c + h + k ) = 2 (c cos  + a )
2 2 2 2 2 2 .....(4)
h
put cos  = in equation (4)
a
h2
we get c2 + h2 + k2 = c2 · 2 + a2
a
c2 a 2
+ + h2a2 k2a2
= c2 h2 + a4
(a2 – c2)h2 + k2a2 = a2(a2 – c2)
h2 k2
+ =1
a2 a 2  c2
x2 y2
 =1 ]
a 2 b2
Home Work after 1st Lecture : Ex-25 (Loney) preferably Q. No. 10 to 22.
 2 LECTURE
5. POSITION OF A POINT RELATIVE TO A PARABOLA :
The point (x1 y1) lies outside , on or inside the parabola y² = 4ax according as the
expression y1²  4ax1 is positive , zero or negative .
6. CHORD JOINING TWO POINTS (t1 and t2) :
The equation of a chord joining t1 & t2 on y2 = 4ax is 2x  (t1 + t2) y + 2 at1 t2 = 0 with
2
slope = t  t
1 2
Note : If the chord joining t1, t2 & t3, t4 pass through a point (c , 0) on the axis, then
t1t2 = t3t4 =  c/a.
EXAMPLES ON 6 .
Ex-1 LOL' and MOM' are two chords of parabola y2 = 4ax with vertex A passing through
a point O on its axis. Prove that the radical axis of the circles described on LL' and
MM' as diameters passes through the vertex of the parabola. [Q.24, Ex-28, Loney]
[Sol. Equation of LL' 2x – (t1 + t2) y + 2at1t2 = 0
passes through (c, 0)
c c
 t1 t2 = – ; |||ly t3 t4 = –
a a
now circle with LL' as diameter
( x  at12 )( x  at 22 ) + (y – 2at1) (y – 2at2) = 0

x2 + y2 – a( t12  t 22 )x – 2a(t1 + 2t)y + c2 – 4ac = 0 ....(1)

(as a 2 t12 t 22  c2 & 4a2t1t2 = 4ac)

|||ly circle on MM' as diameter
x2 + y2 – a ( t 32  t 24 )x – 2a(t3 + t4)y + c2 – 4ac = 0 ....(2)
radical axis i.e. (1) – (2) obviously passes through the origin. ]

Ex.2 A quadrilateral is inscribed in a parabola y2 = 4ax and three of its sides pass through
fixed points on the axis. Show that the fourth side also passes through fixed point on
the axis of the parabola. [T/S, Q.17, Ex-II]
c d
[Sol. Given t1t2 = – ....(1) ; t2t3 = – ....(2)
a a
e
t3t4 = – ....(3)
a
TPT t1t4 = constant
t1 c ec 1
 ....(4) and t1t4 = · = constant]
t3 d d a

Ex-3 A circle and a parabola y2 = 4ax intersect in four points ; show that the algebraic sum
of the ordinates of the four points is zero.
Also show that the line joining one pair of these four points and the line joining the
other pair are equally inclined to the axis. [Q.25, Ex-28, Loney]
[Sol. Let the equation of the circle be
x2 + y2 + 2gx + 2fy + c = 0
solving it with x = at2 , y = 2at
a4 t4 + 4a2 t2 + 2 g a t2 + 4 a f t + c = 0
a4 t4 + 2a (2a + g)t2 + 4 a f t + c = 0
Hence, t1 + t2 + t3 + t4 = 0 ....(1)
or 2a (t1 + t2 + t3 + t4) = 0 Hence proved
2 2
slope of line joining t1 t2 = =– = m1 [using (1)]
t1  t 2 t3  t 4
2
slope of line joining t3 t4 = t  t = m2
3 4
Hence m1 + m2 = 0  Result ]
Ex.4 All chords of the parabola subtending a right angle at the vertex passes through a fixed
point (4a, 0).
Ex.5 Prove that on the axis of any parabola there is a certain point K which has the property
that, if a chord PQ of the parabola be drawn through it, then
1 1
 is the same for all positions of the chord.
PK 2 QK 2
xa y0
[Sol.  r
cos  sin 
x = r cos + c ; y = r sin ; substitute in y2 = 4ax
r2sin2 = 4a(r cos + c)
r2sin2 – (4a cos) r – 4ac = 0
4a cos  4ac
r1 + r 2 = ; r1r2 = –
sin 
2
sin 2 
 16a 2 cos 2  8ac  4
  sin 
r12  r22 (r1  r2 ) 2  2r1 r2  sin 4  2 

1 1  sin 
now,  = = = 2 2
PK 2 QK 2 (r1 r2 )2 (r1 r2 ) 2 16a c

 16a 2 cos2   8ac sin 2  

=   which is a constant if c = 2a and constant = 1 ]
2 2  4a 2
 16a c 
7. LINE / CIRCLE AND A PARABOLA :
y = mx + c & y2 = 4ax
solving them together,
(mx + c)2 – 4ax = 0
m2x2 + 2(cm – 2a)x + c2 = 0 ....(1)

D>0 D=0 D<0

line is a secant Tangent TS

a
now, c = gives the condition of tangency
m
a
hence y = mx + is always a tangent to y2 = 4ax for all m  0.
m
This equation can be used to compute the director circle of the parabola and also to
find the two equations of the tangents to y2 = 4ax which passes through a given point.
 Length of chord of the parabola intercepted on y = mx + c
l2 = (x2 – x1)2 + (y2 – y1)2
= (x2 – x1)2 [1 + m2]
y 2  y1
2 2
= [ (x2 + x1) – 4x1x2 ] [1 + m ] ( x x =m)
2 1

2(cm  2a ) c2
now x1 + x2 = – ; x1 x2 = from (1)
m2 m2
4
hence l = 2
a (1  m 2 )(a  cm)
m
Alternatively ,

1 2 2
l 2 = a2 t 2  t 2  4 a 2 ( t  t ) 2
1 2
2 2 2 2
l = a (t1 – t2) [(t1 + t2) + 4] ....(1)
comparing the equation
2x – (t1 + t2)y + 2at1t2 = 0 with mx – y + c = 0,
2 t t 2 a t1 t 2
we get = 1 2 =
m 1 c
2 c
 t1 + t2 = and t1 t2 =
m am
substituting the values of t1 + t2 and t1 t2 in (1) to get the result. ]
In case length of Focal Chord is required, then
xa y
 r
cos  sin 
x = r cos + a ; y = r sin
hence r2 sin2 = 4a(r cos + a) ; substituting in y2 = 4ax
4a cos  4a 2
we get, r2sin2 – 4a r cos – 4a2 =0 ; r1 + r2 = , r1r2 = –
sin 2  sin 2 
 l2 = | r1 – r2 |2 = (r1 + r2)2 – 4r1 r2

16 a 2 cos2  16 a 2  cos2   sin 2  

16a 2 16 a 2
=  2 =   =
sin  sin   sin 2 
2
sin 4   sin 4 
 | r1 – r2 | = 4a cosec2 ]
EXAMPLES ON 7.
Ex-1 Equation of circle on AB as diameter is
x2 + y2 – (x1 + x2)x – (y1 + y2) y + x1x2 + y1y2= 0.
[Sol. Eliminating x and y from y = mx + c and y2 = 4ax
we get m2x2 + 2 (cm – 2a)x + c2 = 0
2(cm  2a ) x c2
i.e. x2 + + = 0 ....(1)
m2 m2
4ay 4a
|||ly y2 – + c=0 ....(2)
m m
adding (1) and (2) we get the equation of circle with AB as diameter. ]
Ex-2 A circle is described whose centre is the vertex and whose diameter is three quarters of
the latus rectum of a parabola, prove that the common chord of the circle and parabola
bisects the distance between the vertex and the focus. [Q.26, Ex-30, Loney]
3
[Sol. Diameter of the circle = · 4a = 3a
4
9a 2
 equation of the circle is x2 + y2 =
4
solving it with y2 = 4ax
9a 2
x2 + 4ax – =0
4
4x2 + 16ax – 9a2 = 0
4x2 + 18ax – 2ax – 9a2 = 0
2x (2x + 9a) – a(2x + 9a) = 0
(2x – a) (2x + 9a) = 0
a 9a
 x= or x = – (rejected) ]
2 2
8. TANGENTS TO THE PARABOLA y² = 4ax :
(i) Cartesian Tangent : y y1 = 2 a (x + x1) at the point (x1, y1)

y = mx + a (m  0) at  , 
a 2a
(ii) Slope form Tangent :
m  m2 m 
General Note :
(a) Passing through a given point there can be a maximum of two tangents as the quadratic
m2h – km + a = 0 can give atmost 2 real values of m.
k a
This gives m1 + m2 = and m1 m2 = . These relations can be used to compute the
h h
locus of a point from which pairs of tangents to y2 = 4ax enclose an angle .
(b) Director circle of the parabola is x + a = 0 i.e. parabolas own directrix. Hence the set of
points from which parabola can be seen at right angles is x + a = 0.
i.e. parabola's own directrix.
Example : A tangent to the parabola y2 = 8x makes an angle of 45° with the line y = 3x + 5.
Find its equation and also its point of contact.
[Ans. 2x + y + 1 = 0 at (1/2, – 2) ; x – 2y + 8 = 0 at (8, 8)]
[Hint: Let the tangent be y = mx + 2/m ]
1
(iii) Parametric Tangent : t y = x + a t² at (at² , 2at) with slope .
t
General Note : Point of intersection of two tangents at t1 & t2 is at1t2 , a(t1 + t2).
9. NORMALS TO THE PARABOLA y2 = 4ax :
y
(i) Cartesian Normal : y  y1 =  1 (x  x1) at (x1, y1)
2a

(ii) Normal in Slope form : y = mx  2am  am3 at (am2 ,  2am)

General Note :
(a) y = mx + c is normal to y2 = 4ax if c = – 2am – am3. If the parabola is x2 = 4ay then
a 3 1
c = 2a + 2 and if the parabola is y = x2 + 1 then c =  .
m 2 4m 2
(b) If the normal y = mx– 2am – am3 passes through a given point (h, k) then
am3 + (2a – h) m + k = 0
which is a cubic in 'm'. Hence there can be a maximum
of 3 normals to y2 = 4ax which are concurrent at (h, k).
Note that m1 + m2 + m3 = 0 i.e.
(i) algebraic sum of the slopes of 3 concurrent normals is zero.
(ii) algebraic sum of the ordinates of 3 concurrent points is zero.
Example : Prove that the normals at the points, where the straight line lx + my = 1 meets the
 4am 2 4am 
parabola y2 = 4ax, meet on the normal at the point  2 , l  of the parabola.
 l 
[Q.18, Ex-30, Loney]
2
Note : Following lines are normal to y = 4ax
(a) x – y – 3 = 0 (m = 1)
(b) x + y – 3 = 0 (m = – 1)
(c) 2x – y – 12 = 0 (m = 2)
(d) y = (x – 11) cos – cos 3.
(iii)Parametric Normal : y + tx = 2at + at3 at (at2 , 2at) with slope – t.
General Note : Point of intersection of normals at t1 & t2 are, a(t 12 + t 22 + t1t2 + 2),  a t1t2(t1+ t2).

ELEMENTARY EXAMPLES ON TANGENT AND NORMAL (SR. NO. 8 & 9)

LEVEL - 1
Ex-1 If a chord which is normal to the parabola y2 = 4ax at one end subtend a right angle at
the vertex, prove that it is inclined at an angle tan–1 2 to the axis and the normal chord
passes through (4a, 0). [Solved Ex. Pg-184, Loney]
[HInt: y = mx – 2am – am3 ; homogenise ]
Ex-2 Find the equation to the line touching both the
parabolas y2 = 4x and x2 = – 32y.
[Ans. x – 2y + 4 = 0]

Ex-3 Let the tangent to the parabola y2 = 4ax meets the axis in T
and the tangent at the vertex A is y. If the rectangle TAYG is
completed find the locus of G.
[Ans. y2 + ax = 0]

Ex-4 The normal at any point P meets the axis in G and the tangent
at the vertex in Y, if A be the vertex and the rectangle GAYQ
be completed, prove that the equation to the locus of Q is
x3 = 2ax2 + ay2.
[Hint: Start : y + tx = 2at + at3 ] [Q.29, Ex-26, Loney]
Ex-5
(a) Prove that the equation to the circle, which passes
through the focus and touches the parabola y2 = 4ax
at the point (at2, 2at) is
x2 + y2 – ax(3t2 + 1) – ay (3t – t3) + 3a2t2 = 0
If t varies, prove also that the locus of its centre is the curve
27ay2 = (2x – a) (x – 5a)2. [Q.22, Ex-30, Loney]

(b) Through a point P are drawn tangents PQ and PR to a parabola and circles are drawn
through the focus to touch the parabola in Q and R respectively. Prove that the common
chord of these circles passes through the centroid of the triangle PQR.
[Q.24, Ex.30 (Loney)]
2
[Sol. (b) Equation of circle touching the parabola y = 4ax at Q and passing through its
focus is
x2 + y2 – a(3 t12 + 1)x – a(3t2 – t13 )y + 3a2 t12 = 0 ....(1)
|||ly circle touching at Q is
x2 + y2 – a(3 t 22 + 1)x – a(3t2 – t32 )y + 3a2 t 22 = 0 ....(2)
common chord between (1) and (2) is
3(t2 + t1)x + [3 – ( t12 + t 22 + t1t2)] y – 3a(t2 + t1) = 0 ....(3)
Now centroid of the circle PQR is
a 2 2
xg = (t +t + t t )
3 1 2 12
yg = a(t1 + t2)
these two satisfies equation (3). ]

LEVEL - 2
Ex-6 A pair of tangents are drawn which are equally inclined to a straight line y = mx + c
whose inclination to the axis is , prove that the locus of their point of intersection is
the straight line [Q.8, Ex-29, Loney]
y = (x – a) tan 2.
[Sol. We have  = 1 –  =  – 2
 2 = 1 + 2
m1  m 2
tan 2 = tan(1 + 2) = 1  m m
1 2

a
but (h, k) lies on y = mx +  m2h – km + a = 0
m
k a k h k
hence m1 + m2 = and m1 m2 = ; tan 2 = =
h h 1 a h ha
 y = (x – a) tan 2 
Ex-7(a) If tangents are drawn to y = 4ax from any point P on the parabola y = a(x + b) then
show that the normals drawn at their point of contact meet on a fixed line.
[Sol. Let P (at1t2, a(t1+t2) ) ; P must satisfy y2 = a(x + b) [Q.24, Ex-26, Loney]
2 2
Hence a (t1 + t2) = a ( (at1t2) + b) )
a ( t12  t 22  t1t 2 ) = b is true.
Now point of intersection of normals at t1 and t2 are
h = a( t12  t 22  t1t 2 + 2) ....(1)
and k = – at1t2 (t1 + t2) ....(2)
from (1) h = b + 2a  x = b + 2a
(b) Prove that the equation to the circle passing through the points ( at12 , 2at1 ) and

( at 22 , 2at 2 ) and the intersection of the tangents to the parabola at these points is
x2 + y2 – a [(t1 + t2)2 + 2] x – a (t1 + t2)(1 – t1t2) y + a2t1t2(2 – t1t2) = 0.
[Q.27, Ex-28, Loney]
Ex-8 Prove that the two parabolas y2 = 4ax and y2 = 4c (x – b) cannot have a common
b
normal, other than the axis, unless > 2. In other words this gives the condition
a c
for the two curves to have a common normal other than x-axis. [Q.32, Ex-26, Loney]
[Sol. Equation of normal at y2 = 4ax
y = mx – 2am – am3 ....(1)
2
and normal to y = 4c (x – b) with the same slope
y = m (x – b) – 2cm – cm3 ....(2)
If (1) and (2) denotes the same line, we have
– 2am – am3 = – bm – 2cm – cm3
2a + am2 = b + 2c + cm2 (m  0)
(a – c)m2 + 2(a – c) = b
b b
m2 = –2 > 0  >2 ]
a c a c

Ex-9 If a2 > 8b2, prove that a point can be found such that the two tangents from it to the
parabola y2 = 4ax are normals to the parabola x2 = 4by. [Q.33, Ex-26, Loney]
2
[Sol. Equation of tangent to y = 4ax is
a
y = mx + ....(1)
m
Normal to x2 = 4by having slope m is
b
y = mx + 2b + 2 ....(2)
m
comparing (1) and (2)
b a
2b + 2 =
m m
2
2bm – am + b = 0
for two distinct m, D > 0
 a2 – 8b2 > 0  a2 > 8b2 ]
Ex-10 Two equal parabolas have the same focus at the origin and their axes are at right angles,
a normal to one is perpendicular to a normal to the other, prove that the locus of the
point of intersection of these normals is another parabola. [Q.30, Ex-26, Loney]
[Sol. 2 2
x + y = (x + 2a) 2

y2 = 4a2 + 4ax
y2 = 4a (x + a) ....(1)
|||ly equation of the other parabola
x2 = 4a (y + a) ....(2)
Normal at (1) is
y = m(x + a) – 2am – am3
y = mx – am – am3 ....(3)
equation of the normal to parabola (2), in terms of slope is
a
y = m1x + 2a + m 2 ....(4) (calculate)
1
now using m m1 = – 1
my + x = am + am3 ....(5)
equation (3) and (5) passes through (h, k)
k – mh = – am – am3 ....(6)
mk + h = am + am 3 .....(7)
add ——————————
k + h + m (k – h) = 0
hk
 m=
hk
put in equation (6), we get

a (h  k ) a (h  k )
3
(h  k ) h
k– =– –
hk hk ( h  k )3
simplifying, h2 + k2 – 2hk = 2ah + 2ak i.e. x2 + y2 – 2xy – 2ax – 2ay = 0 ]

HOME WORK : Ex-26 (Loney) Q.No. 4, 5, 7, 14, 15, 16, 17, 19, 21, 22, 26, 27, 28, 31, 34

3RD LECTURE
10. THREE SUPPLEMENTRY RESULTS:
(To be derived wherever used in subjective problems)
(a) If t1 & t2 are the ends of a focal chord of the parabola y² = 4ax then t1t2 = 1.
Hence the coordinates at the extremities of a focal chord can be taken as
(at² , 2at) &  a2 , 2a  .
t t 
(b) If the normals to the parabola y² = 4ax at the point t1,
2
meets the parabola again at the point t2, then t2 =  t1  .
t1
Example : Normal to the parabola y2 = 12x at P (3, 6) meet it again at the point Q. Find the
equation of the circle described on PQ as diameter.
[Hint: Q (27, – 18) ]
(c) If the normals to the parabola y² = 4ax at the points t1 & t2 intersect again on the
parabola at the point 't3' then t1 t2 = 2 ; t3 =  (t1 + t2) and the line joining t1 & t2 passes
through a fixed point (  2a, 0).
EXAMPLES :
Ex-1 TP and TQ are tangents to the parabola y2 = 4ax and the
normals at P and Q meet at a point R on the curve, prove
that the centre of the circle circumscribing the triangle TPQ
lies on the parabola [Q.28, Ex-28, Loney]
2
2y = a (x – a).
[Sol. We have t1t2 = 2 ....(1)
now 2k = a(t1 + t2) + 2at3 = a(t1 + t2 + t3) + at3
2k
 t3 =
a
also 2h – 2a + at 32

4k 2 4k 2
2h = 2a + a · 2  2h = 2a +  ha = a2 + 2k2
a a
 2y2 = a(x – a) Henced proved. ]
11. CONCEPT ON ENVELOPE :
If a family of straight lines can be represented by an equation 2P + Q + R = 0 where
 is a parameter and P, Q, R are linear functions of x and y then the family of lines will
be tangent to the curve Q2 = 4 PR.
EXAMPLES :
Ex-1 From the point where any normal to the parabola y2 = 4ax meet the axis is drawn a line
perpendicular to this normal, prove that this line always touches an equal parabola.
[Solved Example, Page 184, Loney] [Ans. y2 = – 4a (x – 2a) ]
12. SUBTANGENT AND SUBNORMAL for y2 = 4ax :
(i) Length of subtangent at any point P(x, y) on the parabola y² = 4ax equals twice the
abscissa of the point P. Note that the subtangent is bisected at the vertex.
(ii) Length of subnormal is constant for all points on the parabola & is equal to the
semi latus rectum.
EXAMPLES :
Ex-1 A parabola is drawn touching the axis of x at the origin and having its vertex at a given
distance k from this axis. Prove that the axis of the parabola is a tangent to the parabola
x2 = – 8k (y – 2k). [Q.38, Ex-26, Loney]
[Sol. Equation of ON
1
y=– x
m
or x + my = 0
solving it with y = mx + c
  cm c 
coordinates of N are  , 
 1  m2 1  m2 
As vertex bisects a subtangent
hence TA = AN
 cm c c
  = 2 and = 2k  c = 2k(1 + m2)
1 m 2 m 1 m 2

Hence equation of the axis of symmetry is

y = mx + 2k(1 + m2)
or 2km2 + mx + 2k – y = 0 ....(1)
 This family of lines touch a curve which is obtained by equating to zero the discriminant
of equation (1).
hence x2 – 8k (2k – y)
or x2 = – 8k (y – 2k) ]
13. PAIR OF TANGENT :
The equation to the pair of tangents which can be drawn from any point
(x1 , y1) to the parabola y2 = 4ax is given by : SS1 = T2 where :
S  y2  4ax ;
S1 = y1  4ax1 ;
2

T  y y1  2a(x + x1).
[Proof : See Article 219 Page-193 (S. L. Loney) ]
14. CHORD OF CONTACT :
(a) Equation to the chord of contact of pair of tangents from (x1 , y1) is
yy1 = 2a (x + x1).
EXAMPLES :
Ex-1 Line 3x + y = 6 intersects the parabola y2 = 4x at A and B.
Find the coordinates of the point of intersection of the
tangents drawn at A and B.

Ex-2 Pair of tangents are drawn to the parabola y2 = – 4x from

every point on the line 3x + y = 2. Show that their chord of
contact passes through a fixed point.
 2 2
[Ans.   ,  ]
 3 3
Ex-3 Two equal parabolas with axes in opposite directions touch
at a point O. From a point P on one of them are drawn
tangents PQ and PQ' to the other. Prove that QQ' will
touch the first parabola in P' where PP' is parallel to the
common tangent at O. [Q.7, Ex-27, Loney]
(b) Length of the chord of contact :
From P (x1, y1) on y2 = 4ax is

y12  4a 2 y12  4ax1

l=
a
Start : x1 = at1 t2
y1 = a(t1 + t2)
l = a2 ( t12  t 22 )2 + 4a2 (t1 – t2)2 = a(t1 – t2)2 [(t1 + t2)2 + 4 ]
= a2 [(t1 + t2)2 – 4t1t2] [(t1 + t2)2 + 4 ]
 y2 4x   y2 
1  1 4
=a  2   
2
1
 a a   a 2 

( y12  4ax1 )( y12  4a 2 ) ( y12  4ax1 )( y12  4a 2 )

l2 =  l= ]
a2 a
(c) Area of the triangle PAB formed by the pair of tangents and their chord of contact
x1 = a t1t2 and y1 = a(t1 + t2)

at12 2at1 1 t ( t  t ) t1  t 2 0
1 a2 1 1 2
now A = at1t 2 a ( t1  t 2 ) 1 = t ( t  t ) t1  t 2 0
2 2 2 12 2
at 22 2at 2 1 t2 2t 2 1

a2 a2
= (t1 – t2)3 = [(t1 – t2)2]3/2
2 2
a2
= [ (t1 + t2)2 – 4t1t2]3/2
2
32 32
a  y1  4x1  a  y1  4ax1 
2 2 2 2
=  2  = 2
2  a a  2  a 
( y12  4ax1 )3 2
A= ]
2a
Ex-1 From a point on the line x + 4a = 0 a pair of tangents are
drawn to the parabola y2 = 4ax. Show that their chord of
contact subtends a right angle at the vertex.
[Sol. Point of intersection of tangents at t1 and t2 is at1t2, a(t1 + t2).
Hence at1t2 = – 4a  t1t2 = – 4
2 2 4 4
now, also mOA · mOB = t · t = t t = =–1 ]
1 2 1 2 4

15. POLAR & POLE :

(i) Equation of the Polar of the point P(x1 , y1) w.r.t. the parabola y² = 4ax is,
y y1= 2a(x + x1)
 n 2am 
(ii) The pole of the line lx + my + n = 0 w.r.t. the parabola y² = 4ax is  , .
1 1 
Note :
(i) The polar of the focus of the parabola is the directrix .
(ii) When the point (x1 , y1) lies without the parabola the equation to its polar is the same
as the equation to the chord of contact of tangents drawn from (x1, y1) when (x1, y1)
is on the parabola the polar is the same as the tangent at the point.
(iv) Two straight lines are said to be conjugated to each other w.r.t. a parabola when the
pole of one lies on the other. Similarly two points P and Q are said to be conjugate
points if polar of P passes through Q and vice versa.
(v) Polar of a given point P w.r.t. any Conic is the locus of the harmonic conjugate of P
w.r.t. the two points is which any line through P cuts the conic.
at 22  at12  2at 2  2at1
Explanation : x1 = ; y1 =
 1  1
at 22  at12  2at 2  2at1
h= ; k=
 1  1
ky1 – 2a(h + x1) = 0
EXAMPLES :
Ex-1 Find the locus of the poles w.r.t. the parabola y2 = 4ax of the tangents to the circle
x2 + 2y = 4a2. [Ans. x2 – y2 = 4a2 ]
[Sol. Let the pole be (h, k)
It's polar w.r.t. y2 = 4ax is
ky = 2a(x + h) ....(1)
now (1) must be tangent to the circle x2 + y2 = 4a2
 2ah
 = 2a
4a 2  k 2
h2 = (4a2 + k2)
 Locus is x2 – y2 = 4a2 ]
Ex-2 If a perpendicular be let fall from any point P upon its
polar w.r.t. y2 = 4ax, prove that the distance of the
foot of this perpendicular from the focus is equal to
the distance of the point P from the directrix.
[Sol. To prove that [Q.5, Ex-27, Loney]
NS = PM
self explanatory ]

Ex-3 Polar of any point P w.r.t. the circle x2 + y2 = a2 touches the parabola y2 = 4ax show
that P lies on the curve y2 = ax.
[Sol. Let P (x1, y1)
its polar w.r.t. the circle x2 + y2 = a2 is
x1 a2
xx1 + yy1 = a2 ....(1) [y=– x ]
y1 y1

a 2 a y
since the tangent to y2 = 4ax   1
y1 x1

y12  ax1  0  locus is y2 = ax. ]

Ex-4 From a variable point on the tangent at the vertex of a parabola y2 = 4ax, a perpendicular
is drawn to its C.O.C. Show that these variable perpendicular lines through a fixed
point on the axis of the parabola. [Ans. (2a, 0) ]
[Sol. chord of contact of P (0, ) is
y = 2ax  2ax – y = 0
line perpendicular to it
x + 2ay = c passes through (0, )  c = 2a
hence equation of PN is
x + 2ay = 2a
2ay = (x – 2a) = 0
This passes through (2a, 0) ]
16. CHORD WITH A GIVEN MIDDLE POINT :
General method for finding the equation of a chord of any conic with middle point (h, k).
Example: Let the parabola be y2 = 8x
and (h, k)  (2, – 3)
we have to find the equation of AB
y + 3 = m(x – 2) ....(1)
now y12 = 4ax1
and y 22 =4ax2
——————
y12 – y 22 = 4a(x1 – x2)
y1  y 2 4a
or x1  x 2 = y1  y 2 = m
but y1 + y2 = 2k = – 6 and 4a = 8
8 4
 m= =–
6 3
4
Hence equation of AB = y + 3 = – (x – 2)  4x + 3y +1 = 0 ]
3
Conversely : To find the mid point of given chord :
Let the equation of the line be 4x + 3y + 1= 0 given
To find the mid point (h, k) of AB
4 4a
here m = – = y  y
3 1 2
4 8
– =  k=–3
3 2k
since 4h + 3k + 1 = 0
4h – 9 + 1= 0  h = 2
hence M is (2, – 3)
For a parabola in particular
Equation of AB
y – k = m (x – h) ....(1)
2
But mAB = t  t
1 2
also 2k = 2a(t1 + t2)
k
(t1 + t2) =
a
2a
 mAB =
k
Hence equation of a chord whose mid point is (h, k)
2a
y–k= (x – h) ....(2)
k
or 4ah + ky – k2 = 2a(x – h) + 4ah = 2a(x + h)
ky  2a ( x  h ) k2
 4
ah
  =  
T S1
EXAMPLES :
Ex-1 Find the locus of the middle point of the chords of the parabola y2 = 4ax which
(i) passes through the focus [Ans. y2 = 2a(x – a) ]
(ii) are normal to the parabola [Ans. y2(y2 – 2ax + 4a2) + 8a4 = 0 ]
(iii) subtend a constant angle  at the vertex (Homogenise)
[Ans. (8a2 + y2 – 2ax)2tan2 = 16a2(4ax – y2)]
(iv) are of given length (say 2l )
(v) are such that the normals at their extremities meet on the parabola
[Ans. y2 = 2a (x + 2a) ; Hint : use t1t2 = 2 ] [Q.19-24, Ex-29, Loney]
[Sol. (i) 2h = a( t12  t 22 ) ....(1)
2k = 2a(t1 + t2) .....(2)
k2 = a2[ t12  t 22 + 2t1t2]

 2h 
k 2 = a2   2 
a 
= 2ah – 2a 2

 y = 2a(x – a)
2

This could also be spelled as locus of the middle point of all focal chords of all the
particles y2 = 4ax. or Locus of the middle point of all the chord of contact of the
pair of tangents drawn from any point on this directrix.

(ii) 2h = a( t12  t 22 ) ....(1) ; k = a (t1 + t2) .....(2)

2 2
also t2 = – t1 –  t1 + t2 = –
t1 t1

k 2 2a
using (2), =–  t1 = –
a t1 k
2a 2k 2a k
 t2 = + +  t2 = +
k 2a k a

2a  2a k  4a 2
 t1t2 = –
k  k  a  = – 2 – k 2 ....(3)

 k2  4 a 2 
from (1) 2h = a [(t1 + t2)2 – 2t1t2] = a  2  2 2  2 
 a  k 

 k2 8a 2   k 4  4a 2k 2  8a 4 
= a 2  4 2  = a  
 a k   a 2k  
2ahk2 = k4 + 4a2k2 + 8a4
k2(k2 – 2ah + 4a2) + 8a4 = 0 Ans. ]
Ex-2 A series of chords is drawn so that their projections on the straight line which is
inclined at an angle  to the axis are of constant length c. Prove that the locus of their
middle point is the curve [Q.15, Ex-28, Loney]
(y – 4ax)(y cos  + 2a sin ) + a c = 0
2 2 2 2
[Sol. Let n̂  cos  î  sin  ˆj

PQ  v  (at 22  at12 ) î  2a ( t 2  t 2 ) ĵ
2h = a ( t12  t 22 ) ; a(t1 + t2) = k

also projection of v on n̂ = c

v · n̂
=c
| n̂ |

a ( t 22  t12 ) cos   2a ( t 2  t 2 ) sin   c

a2(t2 – t1)2 [ a(t2 + t1) cos  + 2a sin  ]2 = c2
a2 [(t2 + t1)2 – 4t1t2] [a(t2 + t1) cos  + 2a sin  ]2 = c2 ]
Ex-3 Through each point of the straight line x = my + h is drawn the chord of the parabola
y2 = 4ax which is bisected at the point. Prove that it always touches the parabola
(y + 2am)2 = 8a(x – h). [Q.25, Ex-29, Loney]
[Sol. Equation of var. chord AB
2a
y – yi = y (x – my1 – h)
i
yyi – yi2 = 2ax – 2amy1 – 2ah
yi2 – (y + 2am)yi + 2a(x – h) = 0
(y + 2am)2 = 8a(x – h) ]
17. DIAMETER :
The locus of the middle points of a system of parallel chords of a Parabola is called a
Diameter. Equation to the diameter of a parabola is y = 2a/m, where m = slope of
parallel chords.
2
Explanation : Slope of AB is m = t  t ....(1)
1 2
k
also 2k = 2a (t1 + t2)  t1 + t2 =
a
2a 2a
 k= ; Hence k =
= constant
m m
2a
Hence equation of the diameter is y = i.e. a line parallel to the axis of the parabola.
m
2a
Solving y = with y2 = 4ax, we have,
m
4a 2 a
= 4ax or x =
m2 m2
 a 2a 
Hence coordinates of Q are  2 , 
m m 
Hence the tangent at the extremity of a diameter of a parabola is parallel to the system
of chords it bisects.
Since point of intersection of the two tangents are A and B is
 2a 
at1t2, a(t1 + t2)  a t1t 2 , 
or
 m
Hence the tangent at the ends of any chords of a parabola meet on the diameter which
bisects the chord.
 Note: A line segment from a point P on the parabola and parallel to the system of
parallel chords is called the ordinate to the diameter bisecting the system of parallel
chords and the chords are called its double ordinate.

18. IMPORTANT HIGHLIGHTS :

(a) If the tangent & normal at any point ‘P’ of the parabola intersect the axis at
T & G then ST = SG = SP where ‘S’ is the focus. In other words the tangent and the
normal at a point P on the parabola are the bisectors of the angle between the focal
radius SP & the perpendicular from P on the directrix. From this we conclude that all
rays emanating from S will become parallel to the axis of the parabola after reflection.
Deduce that, if Q is any point on the tangent and QN is the perpendicular from Q on
focal radius and QL is the perpendicular on the directrix then QL = SN.
[Q.4, Ex-28, Loney]

Note : Circle circumscribing the triangle formed by any tangent normal and x-axis, has
its centre at focus.
(b) The portion of a tangent to a
parabola cut off between the
directrix & the curve subtends a
right angle at the focus.

2at 2t
m1 = = ;
at 2  a t2 1

 a ( t 2  1) ( t 2  1)
m2 = =–
t · 2a 2t

 m1 m2 = – 1 ]
(c) The tangents at the extremities
of a focal chord intersect at
right angles on the directrix,
and hence a circle on any focal
chord as diameter touches the
directrix. Also a circle on any
focal radii of a point P (at2, 2at)
as diameter touches the tangent
at the vertex and intercepts a
chord of length a 1 t 2 on a
normal at the point P.

Note : (1) For computing p draw a perpendicular from S (a, 0) on tangent at P.

(d) Any tangent to a parabola & the perpendicular on it from the focus meet on the tangent
at the vertex.
i.e. locus of the feet of the perpendicular drawn from focus upon a variable tangent is
the tangent drawn to the parabola at its vertex.
Explanation : Tangent 't' is
ty = x + at2 .....(1)
Passing through (h, k) hence
tk = h + at2 .....(A)
A line through (a, 0) with slope – t
y = – t (x – a) ....(2)
(2) also passes through (h, k)
k = – th + at
tk = – t2h + at2 ....(B)
(A) – (B) gives 0 = (1 + t )h2

 x = 0 which is the tangent at the vertex.

(e) If the tangents at P and Q meet in T, then :
 TP and TQ subtend equal angles at the focus S.
 ST2 = SP · SQ &
 The triangles SPT and STQ are similar.
(i) To prove that  = , it will be sufficient to prove that 'T' lies on the angle bisector of
the angle PSQ i.e. perpendicular distance of 'T' from the line SP is equal to the
perpendicular of T from SQ.
equation of SP
2at1
y = at 2  a (x – a)
1

2t1x – ( t12 – 1)y – 2ab1 = 0

2at12 t 2  ( t12  1)a ( t1  t 2 )  2at1

p1 =
( t12  1) 2  4 t12
= a | t1 – t2 |
|||ly p2 = a | t2 – t1 |   = , Hence proved.
Now (ii) SP · SQ = (a + a t12 )(a + a t 22 ) = a2(1 + t12 )(1 + t 22 )
 also (ST)2 = a2(t1t2 – 1)2 + a2(t1 + t2)2 = a2[ t12 t 22 + 1 + t 22 + t 22 ]

= a2(1 + t12 )(1 + t 22 )

Hence (ST)2 = SP · SQ
This is conclusive that product of the focal radii of two points P and Q is equal to the
square of the distance of focus from the point of intersection of the tangents drawn at
P and Q.
ST SQ
(iii) again, = and  = 
SP ST
hence the two triangles SPT and SQT are similar.
(f) Tangents and Normals at the extremities of the latus rectum of a parabola y2 = 4ax
constitute a square, their points of intersection being ( a, 0) & (3a, 0).

[Hint: figure is Self explanatory]

Note :
(1) The two tangents at the extremities of focal chord meet on the foot of the directrix.
(2) Figure L1NL2G is square of side 2 2 a
(g) Semi latus rectum of the parabola y2 = 4ax, is the harmonic mean between segments of
2bc 1 1 1
any focal chord of the parabola is : 2a = i.e.   .
bc b c a

2bc 1 1 1
[Sol. 2a = i.e.  
bc a b c
b = a + at2
 b = a(1 + t2)
a 1
 = ....(1)
b 1  t2
a
c=a+
t2

 1 a t2
 c = a 1  2   = 2 ....(2)
 t  c t 1
from (1) and (2)
a a 1 1 1
+ =1    ]
b c a b c
(h) The circle circumscribing the triangle formed by any three tangents to a parabola passes
through the focus.
TPT  = 
1 1 1
t1 = tan 1 ; t2 = tan 2 ; t 3 = tan3

tan 1  tan 2
 tan  = 1  tan  · tan 
1 2
tan  = | tan(1 – 2) | ....(1)
a ( t 3  t1 ) t 3  t1
m1 = at t  a = t t  1
31 31

1 1

t1 t 3 tan 1  tan 3
= = 1  tan  · tan 
1
1 1 3
t1t 3
m1 = tan (1 + 3)
|||ly m2 = tan (2 + 3)
tan(1  3 )  tan(2  3 )
tan  = 1  tan(   ) · tan(   )
1 3 2 3
tan  = | tan (1 – 2) | ....(2)
from (1) and (2), we get
 =  hence proved ]
(i) The orthocentre of any triangle formed by three tangents to a parabola y2 = 4ax lies on
the directrix & has the coordinates  a , a (t1 + t2 + t3 + t1t2t3).

Find intersection of BD and CE to get 'O'.

(j) The area of the triangle formed by three points on a parabola is twice the area of
the triangle formed by the tangents at these points.
Refer figure of point (h)
at12 2at1 1
at 22 2at 2 1
A1 at 32 2at 3 1
 2
A2 at1t 2 a ( t1  t 2 ) 1
at 2 t 3 a ( t 2  t 3 ) 1
at 3t1 a ( t 3  t1 ) 1

MORE ABOUT NORMALS

If normal drawn to a parabola passes through a point P(h , k) then
k = mh  2am  am3 i.e. am3 + m(2a  h) + k = 0.
2a  h k
Then gives m1 + m2 + m3 = 0 ; m1m2 + m2m3 + m3m1 = ; m1m2m3 =  .
a a
where m1, m2, & m3 are the slopes of the three concurrent normals. Note that the
algebraic sum of the :
 slopes of the three concurrent normals is zero.
 ordinates of the three conormal points on the parabola is zero.
 Centroid of the triangle formed by three conormal points lies on the xaxis.
Example : Consider of the triangle ABC is
a (m12  m 22  m32 )
x1 =
3
2a (m1  m 2  m3 )
y1 = – =0
3
a
now x1 = [(m1 + m2 + m3)2 – 2  m1m 2 ]
3
2 a ( 2a  h ) 2
=– = (h – 2a)
3 a 3
2 2
 centroid is (h – 2a), 0 but (h – 2a) > 0
3 3
 h > 2a
Hence abscissa of the point of concurrency of 3 concurrent normals > 2a.
EXAMPLES :
Ex.1 Find the locus of a point which is such that (a) two of the normals drawn from it to the
parabola are at right angles, (b) the three normals through it cut the axis in points
whose distances from the vertex are in arithmetical progression.
[Ans : (a) y2 = a(h – 3a) ; (b) 27ay2 = 2(x – 2a)3 ] [Ex.237, Pg.212, Loney]
[Sol. (a) we have m1 m2 = – 1
k
also m1 m2 m3 = –
a
k
 m3 =
a
k
put m3 = – is a root of
a
am3 + (2a – h)m + k = 0
 (b) y = mx – 2am – am
hence 2a + am12 , 2a + am22 , 2a + am32
 2 m 22  m12  m 32
2( h  2a )
3 m 22  m12  m 22  m 32 = (m1 + m2 + m3)2 – 2  m1m2  =
a
2(h  2a )
m 22 = which is root of am3 + (2a – h)m + k = 0 ]
3a
Ex.2 If the normals at three points P, Q and R meet in a point O and S be the focus, prove
that SP · SQ · SR = a· SO2. [Ex.238, Pg.213, Loney]
[Sol. SP = a (1  m1 ) ;
2

SQ = a (1  m 2 ) ;
2

SR = a (1  m3 )
2

SP ·SQ ·SR
= (1  m1 ) (1  m 2 ) (1  m3 )
2 2 2
3
a

   
1 

 m  1
2 

 2
  

 2 m1m 2    m1m 2   2m1m 2m3   m1   (m1m 2m3 ) 2

  zero  
Ex.3 A circle circumscribing the triangle formed by three conormal points passes through
the vertex of the parabola and its equation is, 2(x2 + y2)  2(h + 2a)x  ky = 0.
[Q.13, Ex-30, Loney]
[Sol. Equation of the normal at P
y + tx = 2at + at3
passes through (h, k)
at3 + (2a – h)t – k = 0 ....(1)
t1 + t2 + t3 = 0
2a  h k
t1t2 + t2t3 + t3t1 = , t1t2t3 =
a a
Let the circle through PQR is
x2 + y2 + 2gx + 2fy + c = 0
solving circle x = at2, y = 2at
a2t4 + 4a2t2 + 2gat2 + 2f · 2at + c = 0
a2t4 + 2a(2a + g)t2 + 4fat + c = 0 ....(2)
t1 + t2 + t3 + t4 = 0
but t1 + t2 + t3 = 0  t4 = 0  circle passes through the origin
hence the equation of the circle
x2 + y2 + 2gx + 2fy = 0
now equation (2) becomes

at3 + 2(2a + g)t + 4f = 0 ....(3)

 (1) and (3) must have the same root
2(2a + g) = 2a – h
2g = – (h + 2a)
k
and 4f = – k  2f = –
2
Hence the equation of the circle is
k
x2 + y2 – (h + 2a)x – y = 0  2(x2 + y2)  2(h + 2a)x  ky = 0 ]
2

Ex.4 Three normals are drawn to the parabola y2 = 4ax cos from any point on the straight
line y = b sin . Prove that the locus of the orthocentre of the triangle formed by the
x2 y2
corresponding tangent is the ellipse   1 , the angle  being variable.
a2 b2
[Q.15, Ex.30, Loney]
[Sol. y2 = 4Ax where A = a cos 
y + tx = 2At + at3 passes through ,b sin 
b sin  + t  = 2At + At3
At3 + (2A – ) t – b sin  = 0
b sin 
 t1 + t2 + t3 = 0 ; t1 t 2 t 3 =
A
also h = – A = – a cos  ....(1)
and k = A (t1 + t2 + t3 + t1 t2 t3)
b sin 
= A (0 + )
A
k = b sin  ....(2)
from (1) and (2) locus is
x2 y2
 1 ]
a2 b2
LEVEL 3 PROBLEMS
Ex.1 Locus of a point P when the 3 normals drawn from it are such that area of the triangle
formed by their feet is constant. [Q.6, Ex-30, Loney]
[Hint: Area of  ABC = constant

am12  2am1 1
am22  2am2 1 = C
am32  2am3 1

 (m1 – m2)2 (m2–m3)2 (m3–m1)2 = C ....(1)

consider
(m1 – m2)2 = m1  2m1m 2  m 2
2 2

= – [m1 (m2 + m3) + 2m1m2 + m2(m1 + m3)

k
= – [(  m1m 2 ) + 3m1m2] ( m1m2m3 = – )
a
h  2a 3 k
= 
a am3
    
 3k   3k   3k 

(h2a )   (h  2a )   (h  2a )  
 x 3  
m
2  
m
1 
m
 
l  
m   n 

hence equation (1) becomes
(x + l) (x + m) (x + n) = constant
x3 + (l + m + n)x2 + (lm + mn + nl)x + lmn = 0
 1 1 1   1 1 1  27k 3
(h – 2a)3+3k 
m m  2 2   
m3  (h – 2a) +9k  m1m 2 m 2m3 m3m 4  (h –2a)+ m1m 2m3
 1 2
= constant ]

Ex.2 The sides of a triangle touch a parabola y2 = 4ax and two of its angular points lie on
another parabola y2 = 4b(x + c) with its axis in the same direction, prove that the locus
of the third angular point is another parabola. [Q.31, Ex-30, Loney]
[Sol. consider the equations abtained by putting the coordinates of A and B in y2 = 4b(x +c)
a2(t1 + t2)2 = 4b(at1t2 + c) ....(1)
a2(t2 + t3)2 = 4b(at2t3 + c) ....(2)
this implies that t1 and t3 are the roots of the equation,
a2(t + t2)2 = 4b(a t t2 + c) having t1 and t3 as its roots
i.e. a2(t2 + 2tt2 + t 22 ) = 4batt2 + 4bc

 a2t2 + 2t(a2t2 – 2abt2) + a2 t 22 – 4bc = 0

a 2 t 2  2abt 2 t 2 ( 2b  a )
 t1 + t 3 = – =
a2 a

a 2 t 22  4bc k t ( 2b  a ) k
and t1 t3 = ; hence = 2  t2 =
a2 a a 2b  a

h at 22  4bc ah  4bc 2 k2 ah  4bc

=   t2  =
a a2 a ( 2b  a ) 2 a
hence locus is ay2 = (2b – a)2 (ax + 4bc) ]

Ex.3 Circles are drawn through the vertex of the parabola to cut the parabola orthogonally at
the other point of intersection. Prove that the locus of the centres of the circles is the
curve, 2y2(2y2 + x2 – 12ax) = ax(3x – 4a)2 [Q.26, Ex.28 (Loney)]

Ex.4 If the normal at P and Q meet on the parabola, prove that the point of intersection of
the tangents at P and Q lies either on a certain straight line, which is parallel to the
tangent at the vertex, or on the curve whose equation is y2(x + 2a) + 4a3 = 0.
[Q.11, Ex.29 (Loney)]
Ex.5 (a) Prove that infinite number of triangles can be constructed in either of the parabolas
y2 = 4ax and x2 = 4by whose sides touch the other parabola.
(b) Prove that the locus of the centre of the circle, which passes through the vertex
of a parabola and through its intersections with a normal chord, is the parabola
2y2 = ax – a2. [Q.25, Ex.30 (Loney)]