Chapter-17

CONIC SECTION

PRACTICE SHEET
1. If the latus rectum of an ellipse is equal to one half its 2 2
minor axis, what is the eccentricity of the ellipse? 11. Consider the ellipse x  y  1 (b > a). Then, which one
2 2
a b
1 3 of the following is correct?
(a) (b)
2 2 (a) Real foci do not exist
3 15 (b) Foci are ( ae, 0)
(c) (d)
4 4 (c) Foci are ( be, 0)
2. P(2,2) is a point on the parabola y2 = 2x and A is its vertex. (d) Foci are (0,  be)
Q is another point on the parabola such that PQ is 12. Consider the parabolas S1  y2  4ax = 0 and S2  y2  4bx
perpendicular to AP. What is the length of PQ? = 0. S2 will contains S1, if
(a) 2 (b) 2 2 (a) a > b > 0
(b) b > a > 0
(c) 4 2 (d) 6 2
(c) a > 0, b < 0 but |b|> a
3. The focal distance of a point on the parabola y2 = 12 x is 4. (d) a< 0, b > 0 but b > |a|
What is the abscissa of the point?
13. Equation of the hyperbola with eccentricity 3/2 and foci at
(a) 1 (b) 1 (2,0) is 5x2  4y2 = k2. What is the value of k?
(c) 2 3 (d) 2 (a) 4 / 3 (b) 3 / 4
4. If (2, 0) is the vertex and the y  axis is the directrix of a (c) (4 / 3 5) (d) (3 / 4 5)
parabola, then where is its focus? 14. What is the eccentricity of an ellipse, if its latusrectum is
(a) (0, 0) (b) (2, 0) equal to one  half of its minor axis?
(c) (4,0) (d) (4, 0) (a) 1/4 (b) 1 / 2
5. Which one of the following points lies outside the ellipse
(x2 / a2) + (y2 / b2) = 1? (c) 3 / 4 (d) 3 / 2
(a) (a, 0) (b) (0, b) 15. What does an equation of the first degree containing one
(c) (a,0) (d) (a, b) arbitrary parameter passing through a fixed point represent?
6. What is the equation of the parabola, whose vertex and (a) Circle (b) Straight Line
focus are on the xaxis at distance a and b from the origin (c) Parabola (d) Ellipse
respectively? (b > a > 0) x 2 y2
16. The ellipse   1 has the same centricity as the
(a) y2 = 8 (b  a) (x  a) 169 25
(b) y2 = 4 (b + a) (x  a) x 2 y2
(c) y2 = 4 (b a) (x  a) ellipse   1 . What is the ratio of a to b?
a 2 b2
(d) y2 = 4 (b  a) (x  a)
7. If the eccentricity and length of latus rectum of a hyperbola 5 13
(a) (b)
13 and 10 units respectively, then what is the length 13 5
are
3 3 7 8
of the transverse axis? (c) (d)
8 7
7 17. The curve y2 =  4ax where, (a > 0) lies in.
(a) unit (b) 12 unit
2 (a) First and fourth quadrants
15 15 (b) First and second quadrants
(c) unit (d) unit (c) Second and third quadrants
2 4
2 2
(d) Third and fourth quadrants
8. In how many points do the ellipse x  y  1 and the 18. What is the sum of focal radii of any point on an ellipse
4 8 equal to?
circle x2 + y2 = 9 intersect? (a) Length of latus rectum
(a) One (b) Two (b) Length of major axis
(c) Four (d) None of the above (c) Length of minor axis
2 2 2 2 (d) Length of semi latus rectum
9. If the foci of the conics x  y  1 and x  y  1
a 2
7 144 81 25 19. What is the locus of points, the difference of whose
were to coincide, then what is the value of a? distances from two points being constant?
(a) 2 (b) 3 (a) Pair of straight lines (b) An ellipse
(c) 4 (d) 16 (c) A hyperbola (d) A parabola
10. Which one of the following is correct? The eccentricity of 20. x 2 y2
Let E be the ellipse   1 and C be the circle x2 + y2
x2 y2 9 4
the conic   1, (  0) = 9. If P = (1,2), then which one of the following is correct?
a2   b2  
(a) Q lies inside C but outside E
(a) Increases with increase in 
(b) Q lies outside both C and E
(b) Decreases with increase in  (c) P lies inside both C and E
(c) Does not change with  (d) P lies inside C but outside E
(d) None of the above

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21. What are the equations of the directrices of the ellipse 25x2 (b) x2 + y2 = a2  b2
+ 16y2 = 400? (b) x2 + y2 = 2 (a2 + b2)
(a) 3x  25 = 0 (b) 3y  25 = 0 (d) x2 + y2 = 2(a2  b2)
(c) x  15 = 0 (d) y  25 = 0 23. If (4, 0) and (4, 0) are the foci of ellipse and the semi-
22.
2 2
A circle is drawn with the two foci of an ellipse x  y  1 minor axis is 3, then the ellipse passes through which one
a 2
b 2 of the following points?
at the end of diameter. What is the equation of the circle? (a) (2, 0) (b) (0, 5)
(a) x2+y2 = a2 + b2 (c) (0, 0) (d) (5, 0)

ANSWER KEY
1. b 2. d 3. a 4. c 5. d 6. d 7. c 8. d 9. c 10. b
11. d 12. b 13. c 14. d 15. b 16. b 17. c 18. b 19. c 20. d
21. b 22. b 23. d

Solutions
Sol.1. (b)  x1 = 1 2
Length of latus rectum of a hyperbola is 2b
2 Sol.4. (c)
Length of latus rectum of an ellipse is 2b a
a Vertex is (2, 0). Since y  axis is thedirectrix of a
where a is the half of the distance between two
parabola. Equation directrix is x = 0. So, axis of
where b is semi minor axis and a is semi  major vertex of the hyperbola.
parabola is x  axis.
2 2b 2 10
axis. As given, 2b  b Let the focus be (a, 0) Latus rectum = 
a a 3
b 1 5a
 2b  a   or, b2  .....(1)
a 2 (a,0) 3
V
We know that eccentricity (0,0) O (2,0) F
In case of hyperbola, b2 = a2 (e2 1)
...(2)
b2 1 3 Putting values of b2 from equation (1) and
e  1  1 
a2 4 2 13 in equation (2),
Sol.2. (d) e
Distance of the vertex of a parabola from 3
Equation of parabola is y2 = 2x, so vertex lies at directrix = its distance from focus 5a  13 
origin So, OV = VF  (2  0)2 = (a  2)2  a 2   1
So, co  ordinates of vertex are A(0, 0)  a2 = 4a  a = 4
3  9 
Let (x1, y1) be the co  ordinates of the point Q  Focus is (4, 0) 5a 4a 2
or, 
 y12  2x1 ....(i) Sol.5. (d) 3 9
And slope of PQ  y1  2
2 2
The equation of ellipse is x  y  1  0  4a 2  15a  0 or a(4  15a)  0
x1  2 a 2
b 2
15
[co  ordinates of P is (2, 2) as given] 2 2 a  0, hence, a 
The point for which x  y  1  0 is outside 4
Also, slope of AP  2  0  1
2 2
a b Length of transverse axis
20 ellipse.
Since, at (a, 0), 1 + 0  1 = 0. It lies on the 15 15
Since, PQ and AP are perpendicular to each 2a  2  
other, hence slope of AP × Slope of PQ = 1 ellipse. 4 2
At (0, b), 0 + 1  1 = 0 Sol.8. (d)
So, 1  y1  2 
 x 2 It lies on the ellipse. The given equation of circle is : x2 + y2 = 9 and
 1  At (a, 0), 1 + 0  1 = 0 2 2
 y1 2 =  x1 + 2 It lies on the ellipse. ellipse is : x  y  1
 x1 + y1 = 4  x1 = 4  y1 At (a, b), 1 + 1 > 0 4 8
Putting value of x1 in equation (i) So, the point (a, b) lies outside the ellipse. From equation is (1) and (2) we get
y12  8  2y1 or y12  2y1  8  0 Sol.6. (d) x2 9  x2
 1
The parabola’s vertex and focus lie on x  axis is
 y1 =  4 and 2 4 8
at points (a, 0) and (b, 0). Vertex and focus lie on  2x2 + 9  x2 = 8  x2 =  1
Hence, co ordinates of point Q are (8, 4)
the x  axis hence, the axis of parabola is x   x is not real
So, required length
axis. Equation of parabola Vertex whose is a Hence, circle and ellipse do not intersect.
PQ  (8  2)2  (4  2) 2 point (x1, y1) then is(y y1)2 = 4k (x  x1) Sol.9. (c)
So, y1 = 0 and x1 = a and k = distance between
 36  36  72  6 2 focus and vertex = (b  a) so the equation is
2 2
The equation of ellipse is given as: x  y  1
Sol.3. (a) (y  0)2 = 4 (b  a). (x  a) i.e., y2 = 4 (b  a ) (x a 2
7
Focal distance of a point (x1, y1) on the parabola  a) 7
is y2 = 4ax is equal to its distance from directrix Eccentricity is given by: e  1 
Sol.7. (c) a2
x+a = 0 is x1 + a.
For y2 = 12x; comparing with y2 = 4ax.
So, x1 + 3 = 4

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Therefore, foci of ellipse are ( ae, 0) ie, If a and b > 0, then graphic representation would 2

be as follows: e1 = 1  25  12  e  1  b
 7  2 2
 a 1  2  169 13 a
 a 
e1 = e2
Now, the equation of given hyperbola is S2 = y2 - 4bx = 0 2
 12  1  b  a  13
x 2 y2 1 x2 y2 S1 = y 2 - 4bx = 0 2
b 5
    1 13 a
144 81 25 144 81 Sol. 17. (c)
25 25 Left hand parabola always lie in second and third
parabola.
12 9
So, a = and b  Sol. 18. (b)
5 5 Equal to length of major axis.
81/ 25 144  81 225 Sol. 19. (c)
e'= 1    We know that the locus of the difference whose
144 / 25 144 144 S2 will contains S1, distances from two points being constant, is a
15 If latusrectum of S2 > latusrectum of S1 hyperbola.
=  4b > 4a Sol. 20. (d)
12
b>a>0 For a point p (1,2)
 12 15  4(1)2 + 9(2)2  36 = 40  36 > 0
Foci of hyperbola are   .  i.e.,  3,0 Sol.13. (c)
 5 12  Given equation of hypb and 12 + 22  9 =5  9 < 0
Since these foci coincides 5x2  4y2 = k2 point p lies outside of E and inside of C.
Sol. 21. (b)
x2 y2
3 = a 1
7   1 25x2 + 16y2 = 400
2
a2 k k2 x 2 y2
5 4  1
16 25
7
3/a= 1 k k
a2 a = and b  e = 3/5
5 2
so directrix is parallel to y axis.
9/a2 =1–7/a2 The eccentricity 3/2and foci at ( 2,0) of
equation of directrix y = ±b/e
16/a2 =1 a = 4 5x2  4y2 = k2
y = ±25/3
Sol.10. (b) Then, e  3 and  ae = 2 Sol. 22. (b)
Equation of the given conic is an equation of 2 2 2
ellipse k 3 4 Foci of an ellipse x  y  1 are (ae, 0) and
x2
y 2  . 2 k 5 2 2
a b
 (x  0) 5 2 3
(ae, 0) equation of circle with centre (0,0) and
a 2   b2   Sol.14. (d)
 A2 = a2 +  and B2 = b2 +  radius ae is x2 + y2 = (ae)2 [where, (ae)2 = a2  b2]
2
Since, Latusrectum of an ellipse = 2b x2 + y2 = a2  b2
Eccentricity, e  1  B  1  b  
2 2
a Sol. 23. (d)
A 2
a 
2
and mnor axis = 2b 2ae = 8  ae = 4
a 2    b2   a 2  b2 2b2 We know that
  b   a  2b 2
a 
2
a2   b2 4  9 
a e= 1     1  2 
 is in the denominator so, when  increases, the a2 a  a 
b2 b23 3
Also,e  1   1  
eccentricity decreases.  16  a  9  a2 = 25  a = 5
2
2
a 4b2 4 2
Sol.11. (d) a2 a2
2 2 Sol.15. (b)
Given equation x  y  1
2 2

2 2
From the given information, we have an equation Thus, the equation of the ellipse is a  y  1
a b of the first degree which contains one arbitrary 25 9
Since b > a parameter. Therefore the required equation Which is satisfied by (5, 0). Hence, the ellipse
 Foci = (0,  be) represents a straight line. passes through (5, 0).
Sol.12. (b) Sol. 16. (b)

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 NDA PYQ
1. What are the points of intersection of the curve 4x2  9y2 =  
(c)  0,  13  (d) None of these
1 with its conjugate axis?  6 

(a)  1 , 0  and   1 , 0  [NDA (II) - 2013]
2   2  11. The axis of the parabola y2 + 2x = 0 is:
(b) (0, 2) and (0,2) (a) x = 0 (b) y = 0
(c) (0, 3) and (0, 3) (c) x = 2 (d) y = 2
(d) No such points exist [NDA (II) - 2013]
[NDA (I) - 2011] 12. The length of latus rectum of the ellipse 4x2 + 9y2 = 36 is:
2. What is the sum of the focal distances of a point of an (a) 4/3 (b) 8/3
2 2
ellipse: x  y  1 (c) 6 (d) 12
a2 b2 [NDA-2013(2)]
(a) a (b) b 13. What is the equation of parabola whose vertex is at (0,0)
(c) 2a (d) 2b and focus is at (0, 2)?
[NDA-2011(1)] (a) y2 + 8x = 0 (b) y2  8x = 0
3. What is the focal distance of any point P(x1,y1) on the (c) x2 + 8y = 0 (d) x2  8y = 0
parabola y2 = 4ax? [NDA (I) - 2014]
(a) x1 + y1 (b) x1y1 14. What is the sum of the major and minor axes of the ellipse
(c) ax1 (d) a + x1 whose eccentricity is 4/5 and length of latus rectum is 14.4
[NDA-2011(2)] units?
(a) 32 units (b) 48 units
4. What is the eccentricity of the conic 4x2 + 9y2 = 144?
(c) 64 units (d) None of these
5
(a) 5 (b) [NDA (I) - 2014]
3 4 15. What is the length of the latus rectum of an ellipse 25x2 +
16y2 = 400?
(c) 3 (d) 2
3 (a) 25/2 (b) 25/4
5
(c) 16/5 (d) 32/5
[NDA (I) - 2012]
[NDA (II) - 2014]
5. If the latus rectum of an ellipse is equal to half of the minor 16. The point on the parabola y2 = 4ax nearest to the focus has
axis, then what is its eccentricity? its abscissa.
(a) 2/3 (b)1/3 (a) x = 0 (b) x = a
(c) 3/2 (d) 1/2 (c) x = a/2 (d) x = 2a
[NDA (I)-2012] [NDA (I) - 2015]
6. The sum of the focal distances of a point on the ellipse 2 2
17. The hyperbola x  y  1 passes through the point (3 5,1 )
x 2 y 2 =1 is: 2
a2
b

4 9 and the length of its latus rectum is 4/3 units. The length of
(a) 4 units (b) 6 units the conjugate axis is:
(c) 8 units (d) 10 units (a) 2 units (b) 3 units
[NDA (II) - 2012] (c) 4 units (d) 5 units
7. The eccentricity e of an ellipse satisfies the condition. [NDA (I) - 2015]
(a) e < 0 (b) 0< e < 1 2 2
18. Consider any point P on the ellipse x  y =1 in the first
(c) e = 1 (d) e > 1 25 9
[NDA (II) - 2012] quadrant. Let r and s represent its distances from (4,0) and
8. The difference of focal distances of any point on a (4,0) respectively, then (r + s) is equal to:
hyperbola is equal to: (a) 10 unit (b) 9 unit
(a) Latus rectum (b) Semi-transverse axis (c) 8 unit (d) 6 unit
(c) Transverse axis (d) Semi-latus rectum [NDA (II) - 2015]
[NDA (I) - 2013]
19. The eccentricity of the hyperbola 16x2  9y2 = 1 is?
9. The equation of the ellipse whose vertices are at (±5,0) and
(a) 3/5 (b) 5/3
foci at (±4,0) is:
(c) 4/5 (d) 5/4
2 2 2 2
(a) x  y  1 (b) x  y  1 [NDA (II) - 2015]
25 9 9 25 20. The equation of the hyperbola having latus rectum and
2 2 2 2
(c) x  y  1 (d) x  y  1 eccentricity 8 and 3 respectively?
16 25 25 16 5
[NDA-2013(1)] x 2 y2 x 2 y2
10. The foci of the hyperbola 4x2  9y2 1 = 0 are: (a)  1 (b)  1
25 20 40 20

(a)  13,0  (b)   13 , 0 
  x 2 y2 x 2 y2
 6  (c)  1 (d)  1
40 30 30 25

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 [NDA (II) - 2016] [NDA (II) - 2017]
21. What is the eccentricity of rectangular hyperbola? 30. What is the equation of the ellipse whose vertices are
(a) 2 (b) 3 (, 5, 0) and foci are at (, 4, 0)?
2 2 2 2
(c) 5 (d) 6 (a) x  y  1 (b) x  y  1
[NDA (II) - 2016] 25 9 16 9
2 2
22. If the ellipse 9x2 + 16y2 = 144 intercepts the line 3x + 4y = 2 2
(c) x  y  1 (d) x  y  1
12, then what is the length of the chord so formed? 25 16 9 25
(a) 5units (b) 6units [NDA - (I) 2018]
(c) 8units (d) 10units 31. Equation 2x2 - 3y2 -6 = 0 represents
[NDA (II) - 2016] (a) A circle (b) A parabola
Direction (for next two): Consider the following for the (c) An ellipse (d) A hyperbola
next two items that follow: [NDA - (I) 2019]
Consider the parabola y = x2 + 7x + 2 and the straight line y 32. Two parabolas y2 = 4ax and x2 = 4ay intersect at
= 3x 3. (a) at two points located on the line y = x
23. What are the coordinates of the point on the parabola which (b) only at origin
is closest to the straight line? (c) At three points out of them one is lies on line y + x = 0
(a) (0, 2) (b) (2, 8) (d) only at (4a,4a)
(c) (7, 2) (d) (1, 10) [NDA - (I) 2019]
[NDA (II) - 2016] 33. The sum of the focal distances of a point on an ellipse is
24. What is the shortest distance from the above point on the constant and equal to:
parabola to the line? (a) length of minor axis
10 (b) length of major axis
(a) (b) 10 (c) length of letus rectum
2 5
(d) sum of lengths of minor axis and major axis
(c) 1 (d) 5 [NDA - (I) 2019]
10 4 34. If the angle between the lines joining the end points of the
[NDA (II) - 2016] x 2 y2 
25. What is the equation of the ellipse having foci (2, 0) and ellipse 2  2  1 with one of its foci is then what is
a b 2
the eccentricity 1/4?
2 2 the eccentricity of the ellipse?
(b) x  y  1
2 2
(a) x  y  1 1
64 60 60 64 (a) (b) 1
2 2 2 2
(d) x  y  1
2 2
(c) x  y  1
20 24 24 20 3 1
(c) (d)
[NDA (I) - 2017] 2 2 2
26. The position of the point (1,2) relative to the ellipse 2x2 + [NDA - (II) 2019]
7y2 = 20 is: 35. Let P(x,y) be any point on ellipse 25x2 + 16y2 = 400. If
(a) Outside the ellipse Q(0,3) and R (0,−3) are two points, then what is (PQ + PR)
(b) Inside the ellipse but not at the focus equal to ?
(c) On the ellipse (a) 12 (b) 10
(d) At the focus (c) 8 (d) 6
[NDA (II) - 2017] [NDA 2020]
27. The equation of the ellipse whose centre is at origin, major 36. In the parabola, y2 = x, what is the length of the chord
3 passing through the vertex and inclined to the x-axis at an
axis is along x-axis with eccentricity and latus rectum 4
4 angle θ?
units is: (a) sinθ.sec2θ (b) cosθ.cosec2θ
2 2
49x 2 7y 2 (c) cotθ.sec θ
2 (d) 2tanθ.cosec2θ
(a) x  7y  1 (b)  1 [NDA 2020]
1024 64 1024 64 37. If any point on a hyperbola is (3tan, 2 sec), then what is
2 2
x2 y2 the eccentricity of the hyperbola?
(c) 7x  49y  1 (d)  1
1024 64 1024 64 3 5
(a) (b)
[NDA (II) - 2017] 2 2
28. Geometrically Re (z2  i) = 2, where i = 1 and Re is the
(c) 11 (d) 13
real part, represents:
2 2
(a) Circle (b) Ellipse
(c) Rectangular hyperbola (d) Parabola [NDA (I) 2021]
[NDA (II) - 2017] 38. Consider the following with regard to eccentricity (e) of a
29. A man running round a racecourse notes that the sum of the conic section:
distance of two flag-posts from him is always 10 m and the 1.e = 0 for circle 2.e =1 for parabola
distance between the flag posts is 8m. The area of the path 3. e < 1 for ellipse
he encloses is: Which of the above are correct?
(a) 18sq m (b) 15sq m (a) 1 and 2 only (b) 2 and 3 only
(c) 12sq m (d) 8sq m

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 (c) 1 and 3 only (d) 1, 2 and 3 (a) 1 and 2 only (b) 2 and 3 only
[NDA-(I) 2021] (c) 1 and 3 only (d) 1, 2 and 3
Consider the following for next (02) question. [NDA – 2023 (1)]
The two ends of letus rectum of a parabola are (–2,4) and 47. What is the equation of directrix of parabola y2 = 4bx,
(–2,–4). where b < 0 and b2 + b – 2 = 0?
39. What is the maximum number of parabolas that can be (a) x+1 = 0 (b) x–2=0
drawn through these two points as end points of latus (c) x–1=0 (d) x+2=0
rectum? [NDA-2023 (2)]
(a) Only one (b) Two 48. Consider the following in respect of the equation
(c) Four (d) Infinite x2 y2
[NDA-(II) 2021]  2
24  k k  16
40. Consider the following statements in respect of such
1.The equation represents an ellipse if k=19
parabolas:
2.The equation represents an hyperbola if k=12
1.One of the parabolas passes through the origin (0, 0)
3.The equation represents an circle if k=20
2.The focus of one of the parabolas lies at (–2, 0)
How many of the statements given above are correct?
Which of the above statements is/are correct?
(a) only one (b) only two
(a) 1 only (b) 2 only
(c) all three (d) none
(c) Both 1 and 2 (d) Neither 1 nor 2
[NDA-2023 (2)]
[NDA-(II) 2021] 49. Consider the following statements in respect of hyperbola
41. What is the equation of the parabola with focus (–3, 0) and
direction x – 3 = 0? x2 y2
 =1.
(a) y2 = 3x (b) x2 = 12 y cos2  sin 2 
(c) y2 = 12x (d) y2 = – 12x 1.The two foci are independent of 
[NDA-(I) 2022] 2.The eccentricity is sec
42. What is the distance between the foci of the ellipse x2 + 2y2 3.The distance between the two foci is 2 units.
= 1? How many of the statements given above are correct?
(a)1 (b) √2 (a) only one (b) only two
(c) 2 (d) 2√2 (c) all three (d) none
[NDA-(I) 2022] [NDA-2023 (2)]
43. The centre of an ellipse is at (0, 0), major axis is on the y- 50. In the parabola y2 = 8x, the focal distance of a point P lying
axis. If the ellipse passes through (3, 2) and (1, 6), then on it is 8 units. Which of the following statements is/are
what is its eccentricity? correct?
(a) 3 (b) √3 
1. The coordinates of P can be 6,4 3 . 
2 2. The perpendicular distance of P from the directrix of
(c) 5 (d) 5 parabola is 8 units
2 Select the correct answer using the code given below:
[NDA 2022 (II)] (a) 1 only (b) 2 only
44. An equilateral triangle is inscribed in a parabola x2 = √3𝑦 (c) Both 1 and 2 (d) Neither 1 nor 2
where one vertex of the triangle is at the vertex of the [NDA-2024 (1)]
parabola. If p is the length of side of the triangle and q is 51. What is the eccentricity of the ellipse if the angle between
the length of the latus rectum, then which one of the the straight lines joining the foci to an extremity of the
following is correct? minor axis is 90°?
(a) p = q (b) p = √3𝑞 (a) 1 (b) 1
(c) p = 2√3𝑞 (d) 2√3𝑝 = 𝑞 3 2
[NDA 2022 (II)] (c) 1 (d)
1
Consider the following for the next two (02) items that 3 2
follow: [NDA-2024 (1)]
P(x, y) is any point on the ellipse x2 + 4y2 = 1. Let E, F be 52. The foci of the ellipse 4x2 + 9y2 = 1 are Q and R. If P(x,y)
the foci of the ellipse. is any point on the ellipse, then what is PQ + PR equal to ?
45. What is PE + PF equal to? (a) 2 (b) 1
(a) 1 (b) 2 (c) 2/3 (d) 1/3
(c) 3 (d) 4 [NDA-2024 (2)]
[NDA – 2023 (1)] 53. Consider the points P(4k,4k) and Q(4k, – 4k) lying on the
46. Consider the following points: parabola y2 = 4kx. If the vertex is A, then what is ∠PAQ
 
1.  3 ,0  2.  3 , 1  equal to ?
 2   2 4 (a) 60° (b) 90°
   
(c) 120° (d) 135°
 3 1
3.  ,  [NDA-2024 (2)]
 2 4 

Which of the above points lie on latus rectum of ellipse

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 ANSWER KEY

1. d 2. c 3. d 4. a 5. c 6. b 7. b 8. c 9. a 10. b
11. b 12. b 13. c 14. c 15. d 16. a 17. c 18. a 19. b 20. a
21. a 22 a 23. b 24. c 25. a 26. a 27. b 28. c 29. b 30. a
31. d 32. a 33. b 34. b 35. b 36. b 37. d 38. a 39. b 40. c
41. d 42. b 43. a 44. c 45. b 46. d 47. b 48. c 49. c 50. c
51. d 52. b 53. b

Solutions
=   1  1 , 0     13 , 0 
Sol. 1. (d) Putting the value of a in Eq. (ii), we get
A hyperbola never meet/intersect conjugate axis     b2 = 7.2  20
in real points.  4 9   6  b2 = 72  2 = 144  b2 = (12)2
Sol. 2. (c) Sol. 11. (b) Hence, the sum of the major and minor axes
Sum of focal distances is always equal to length Given equation of parabola is: = 2a + 2b
of major axis = 2a y2 + 2x = 0  y2 =  2x =2(a+b) = 2 (20+12)
Sol. 3. (d) Which is of the form y2 =  4ax =2  32 = 64 units
Focal distance is always equal to distance of that So, axis of the parabola is y = 0 Sol. 15. (d)
point form directrix i.e. a + x1 Sol. 12. (b) Equation of ellipse is 25x2 + 16y2 = 400
Sol. 4. (a) length of latus rectum of the ellipse 4x2 + 9y2 = 2 2
36  x  y 1
The given equation of conic is 16 25
2 2 x2/9 + y2/4 = 0
4x2 + 9y2 = 144  x  y  1 a = 3, b = 2 Here, a2 = 16 and b2 = 25
36 16 LR = 8/3 length of latus rectum
Which represent an ellipse, here a > b
= 2a  2 16  32
2
Sol. 13. (c)
a2 = 36  a = 6 Given vertex of the parabola = (0, 0) b 5 5
b2 = 16 b=4 and focus of the parabola = (0, 2) Sol. 16. (a)
Now, eccentricity, b2 = a2 (1e2) Let P be any point on the parabola, then equation Required abscissa is x = 0
16 = 36 (1e2)  4 =  e2 directrix is y  2 = 0 Sol. 17. (c)
Equation of parabola is
9
 PS  PM | y 2 |
Since, hyperbola passes through 3 5,1  
e2 = 1  4  5 e= 5 
3 5 
2
 x  0   y  2
2 2
9 9 3 1 1
=2  2 1
 
Sol. 5. (b)  2
= |y2|2
a2 b
x2   y  2
2
Since, the sum of foci distances of a point on the
2 2
x2 + y2 + 4 + 4y = y2 + 4  4y  45  1  1
ellipse x  y  1 is equal to 2b.
a 2 b2
a 2
b 2
x2 =  8y, which is the required equation of
When b > a parabola. 45b2  a2 = a2 b2 …(i)
2
a2 = 4, b2 = 9 a = 2, b = 3 Sol. 14. (c) Also, 2b  4
Sum of the focal distances = 2  3 = 6 units We know that, length of major axes of an ellipse a 3
Sol. 6. (b) = 2a and length of minor axes of an ellipse = 2b 6b2 = 4a
The eccentricity of ellipse lies between 0 and 1 Given that, 2

eccentricity of an ellipse =4/5…(i) a = 6b …(ii)

Sol. 7. (b) 4
For ellipse 0 < e < 1 and length of latus rectum of an ellipse = 14.4
On putting the value fro Eq. (ii) in Eq. (i), we get
units
Sol. 8. (c) 2 2
45b2   6b    6b  b 2
2 2
2 2
By definition of hyperbola,  2b  14.4  b  7.2    
A hyperbola is the set of points in a plane, where a a  4   4 
b2 = 7.2a …(ii)
2 4 2
distances from two fixed points in the plane have 45b2  36b  36b .b
a constant difference i.e., transverse axis. The Since, 16 16
two fixed points are the foci of the hyperbola. b2 = a2 (1e2)
45b2 = 6b  b 2  1
2

Sol. 9.  
7.2 a = a2 1   4  
2
16
   
  5   45  16 = 36b2 (b2 + 1)
Sol. 10. (b)
Given equation of hyperbola is [from eqs.(i) and (ii)] b4 + b2 = 20
b=2
2
4x2  9y2 = 1  x  y  1
2
7.2a = a2  1  16  2b = 4 = length of conjugate axis
 
   
1/4 1/9  25 
Sol. 18. (a)
Here: a2 =1/4 and b2 =1/9 7.2a = a2 = 9
Foci of the hyperbola = ( ae, 0) 25
  9a2  7.2  25a = 0

=  a a  b ,0    a 2  b2 ,0

2 2
9a2  36  5a = 0
 a 
  9a(a20) = 0
 a = 20 (a≠0)

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 Y
e= 3 …(i)
Q (0,3) 4
P
b2 3 a 2  b2 9
3 (4,0) 1 2
  
X X  a 4 a2 16
(4,0) P
(4,0)
a 2  2a 9
   a2  9
a2 16 a 16
Y
7a = 32  a =
32
The sum of the distance of any point P from the Given ellipse intercepts the line at point (4,0) and
(0,3).
7
foci of an ellipse
length of chord b2 = 2  32  64
 ae   b2
2
=2 7 7
 4  0   0  3
2 2
PQ =
= 2 16  9 Now, required equation of ellipse is given as:
=5 unit x2 y2
r + s = 10cm  1
Solutions (for next two): 2
 64 
Sol. 19. (b)  32 
2
   
Equation of Hyperbola: 16x2  9y2 = 1 Given, y = x2 + 7x + 2 =  x  7   41  7   7 
 
a2 = 1 and b2 = 1  2 4 2 2

Let the co-ordinates of the point on this parabola or 49x  7y  1

16 9 1024 64
2
be
e = 1  b  1  16  5  7 2 41  Sol. 28. (c)
a2 9 3 P  ,P   Re (z2  i) = 2 or Re [(x+iy)2  i] = 2 {where, z
 2 4
= x + iy} or Re (x2  y2 + 2ixy  i) = 2 or x2  y2
Sol. 20. (a) The equation on the line is y = 3x  3  y  3x +
Given, length of latus rectum =2
3=0
2
= 2b  8
x 2 y2
…(i) Distance of the point from the line
or  1
a  2 41   7 2 2
 P    3 P    3
b 3
2 z=  4  2 Sol. 29. (b)
and eccentricity = 1    10 P
a 5
2  dz  1  2p  3  0
1+  b   9 dt 10 F2 F1
 
a 5 P=3/2
P( PF1 + PF2 = 10
Sol. 23. (b)
2a = 10  a = 5
2,4

Co-ordinates of the point on the parabola which

F1F2 = 8 2c = 8  c= 4
)

is closest to the straight line

a2 = b 2 + c2  b 2 = 3 2  b = 3
=  P  7 , P 2  41    3  7 ,  3   41 
2

     Area = ab =   3  5 = 15 sq.m

4 
0
(1,3)  2 4  2 2 2 Sol. 30. (a)
=(2, 8)  For P  3  c=4, a = 5, b2 = 25  16 = 9
 2 
2 2


b2 4  Equation of ellipse is x  y  1
 …(ii)
25 9
a2 5 Sol. 24. (c)
On solving equations (i) and (ii), we get Shortest distance from the point to parabola Sol. 31. (d)
2
a2 = 25 and b2 = 20 41  7  3  41  3 7  x 2 y2
P2   3 P    3     3    3  1
Equation of hyperbola will be: = 4  2
 
2 4 2 2 3 2
x 2 y2 10 10
 1 9 41
this equation represents a hyperbola
 63
= 8  6  3  1 unit
25 20 Sol. 32. (a)
=4 4
Sol. 21. (a) x2 = 4ay and y2 = 4ax.
10 10 10
We know that, by solving these equations solutions are
Equation of rectangular hyperbola Sol. 25. (a) (0,0) and (4a,4a) these points are lie on y = x line
2 2 The foci of the ellipse is (2, 0) Sol. 33. (b)
x  y = p  x  y 1
2 2 2
i.e.  ae =  2
p2 p2 Sum of focal distances is always equal length of
and e=1/4 major axis.
Here, length of transverse axis and conjugate a = 2  4 = 4 Sol. 34. (b)
axis are equal. and b2 = a2 (1e2) = 64
2
If F is focus of ellipse and F( ae, 0)
Eccentricity = 1   b   1   p   1  =60
2
A is one end point of minor axis on positive y
    1  
a p  16  axis A(0,b)
2 2 B is second end point of minor axis on negative y
= 11  2 the required equation is x  y  1 axis B(0,−b)
Sol. 22. (a) 64 60
AF and BF lines are given perpendicular.
Given, Sol. 26. (a) So slope of AF × slope of BF = −1
9x2 + 16y2 = 144 Equation of ellipse: 2x2 + 7y2 = 20
 b  0   b  0 
2 2
 x  y  1 and 3x + 4y = 12
Putting x = 1 and y = 2 on LHS, we get     1
2(1)2 + 7(2)2 = 2 + 28 = 30 > 20  0  ae   0  ae 
16 9
Sol. 27. (b)  b 2 
 x  y 1 …(ii) Latus rectum = 4   2 2   1
4 3 a e 
2b 2
From figure, or 4 b 2 eccentricity
a  e2 
2
or b = 2a a2

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 b2 e= 1

1 x2 y2
e  1  1  e2  2
a2 2 2 24  k k  16
distance between foci
⇒ e2 = 1 − e2 If k = 19
2ae = 2  1  1  2
⇒ 2e2 = 1 x 2 y2
2  2
1 5 3
e Sol. 43. (a)
2 2 2
x 2 y2
equation of ellipse x  y  1 where(b > a)  1
Sol. 35. (b) 2 2
10 6
a b
Given ellipse 25x2 + 16y2 = 400
ellipse passes through (3, 2) and (1, 6) Equation represents ellipse.
x2 y 2
 1 9

4
1 …….(i) If k = 12
16 25 a2 b2
a = 4 , b = 5 end points of major axes are (0,5) x 2 y2
1 16 …….(ii)  2
and (0,−5) coordinates of focii are Q(0,3) and R  1 12 4
(0,−3) a2 b2
(PQ+PR)=sum of focal distances = length of by both equations a2 = 35 and b2 = 140 Equation represents hyperbola.
major axes = 10 unit a2 35 3 If k = 20
e  1  1 
Sol. 36. (b) b2 140 2 x 2 y2
chord inclined with x axis at angle θ  2
Sol. 44. (c) 4 4
and passing through vertex (origin)
then equation of chord will be y = tanθ.x x2 + y2 = 8
it will cut the parabola at A(0,0) and equation represents circle.
B(cot2θ,cotθ) All three statements are correct
length of chord = cot 4   cot 2  Sol. 49. (c)
= cot  cot 2   1 x2 y2
 1
= cot  cosec cos 2  sin 2 
= cosθ.cosec2θ
Sol. 37. (d) sin 2 
e  1  sec 
any point on a hyperbola is (3tan, 2 sec), cos 2 
x = 3tan, ⇒ tan = x/3
y = 2sec ⇒ sec = y/2 Focii are ae = cos.sec = 1
sec2 −tan2 = 1 side of triangle is p , so its horizontal component Distance between (1,0) and (-1,0) is 2 units.
is pcos60 and vertical component is psin60.
y 2 x2 Sol. 50. (c)
 1 so coordinates of A is  p 3 p 
4 9  , 
2 2 
9 13 13  
e  1  
4 4 2 this point lie on parabola so this will satisfy
Sol. 38. (a) equation of parabola x2 = √3y
for ellipse 0 < e < 1 p2 3p
 3
Sol. 39. (b) 4 2
End point of LR are (–2, 4) and (–2, –4) p=6
Length of LR = 8 unit and letus rectum = q = √3
Focal length = 2 unit i.e. p/q = 6/√3
Two parabola are possible p = 2√3q
Sol. 45. (b)
ellipse x2 + 4y2 = 1
a = 1 and b = ½
sum of focal distances is equal to length of major
axis = 2a = 2 according to above diagram
a + x1 = focal distance (for all y2 = 4ax)
Sol. 46. (d) 2 + x1 = 8 (here a = 2 is focal length)
ellipse x2 + 4y2 = 1
x1 = 6
a = 1 and b = ½
1 3

put x = 6 in equation of parabola so P is 6,4 3

e  1  a + x1 = perpendicular distance from directrix.
4 2
both statements are correct.
 3  Sol. 51. (d)
focii (ae,0) =  ,0  coordinates of foci F1 (ae,0) and F2(–ae,0)
 2 
  one end point of minor axis is A(0,b)
Sol. 40. (c) angle between AF1 and AF2 is 90°
3
From above graph statements are correct. and L.R. equation x  we know that if two lines are perpendicular than
Sol. 41. (d) 2 m1m2 = –1
focus (–3,0) slope of AF1 = 0  b   b
All given points are lie on this line
director x–3 = 0 ae  0 ae
parabola y2 = – 12x Sol. 47. (b) slope of AF2 = 0  b  b
Sol. 42. b2 + b – 2 = 0
 ae  0 ae
b = – 2 given (b<0)
x2 y2  b  b 
 1 equation of directrix of parabola y2 = -8x      1
1 1/ 2 equation of directrix is x = 2 and x = - b  ae  ae 
Sol. 48. (c) b2
 e2
a2

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and eccentricity of ellipse Sol. 52. (b) so length of major axis = 2a =1
b2 4x 2  9 y 2  1 Sol. 53. (b)
e  1  2  1  e2 P(4k,4k) , Q(4k,4k) , A(0,0)
a x2 y2
 1 slope of PA = 1 and PQ = – 1
e2  1  e2 1/ 4 1/ 9 here m1m2 = – 1 so both lines are perpendicular.
1 Sum of distances of any point on ellipse from
2e 2  1  e  both axes = length of major axis
2 here a = 1/2

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