Conic Sections
Conics – comes from the word cone which is where the shape of parabolas, circles,
ellipses, and hyperbolas originate.
How to recognize the type of conic:
Two coefficients to examine are A and C.
• For circles, the coefficients of x2 and y2 are the same and the same value: A = C
• For ellipses, the coefficients of x2 and y2 are the same sign and different values: A,C
> 0, A≠C
• For parabolas, either the coefficient of x2 or y2 must be zero: A = 0 or C = 0
• For hyperbolas, the coefficients of x2 and y2 are opposite signs: C < 0 < A or A < 0 <
C
What type of conic is each of the following relations?
1. 5𝑦 2 − 2𝑥 2 = −25 1. 7x2 − 5y2 = 35
1
2. x = 3y2 – 4
1
2. 𝑥 = − 2𝑦2 − 3 3. 9x2 + 4y2 =36
1
4. x2 − 5y = -2
3. 4𝑥 2 + 6𝑦 2 = 36 𝑥2 𝑦2
5. − 9 =1
1 4
4. 𝑥 2 − 4𝑦 = 1 6. 10x2 − 6y2 = 60
7. y = −2x2 + 7x + 1
𝑥2 𝑦2 8. 2x2 + 3y2 = 12
5. − + =1
8 4 9. x2 + 16y = 8
𝑥2 𝑦2
6. −𝑥 2 + 99𝑦 2 = 12 10. 25 + 16 = 1
11. −x2 + 4y2 =20
12. y2 = −4x + 9 y2=−4x+9
1. 3𝑥 2 = 3𝑦 2 + 18 1. 3x 2 + 4y 2 = 12
2. 𝑦 = 4(𝑥 − 3)2 + 2 2. x 2 + y 2 = 9
3. 𝑥 2 + 𝑦 2 = 4 3. x 2 /4 + y 2 /9 = 1
4. 𝑦 2 + 2𝑦 + 𝑥 2 − 6𝑥 = 12
4. y 2 + x = 11
𝑥2 𝑦2
5. 6
+ 12 = 1 5. x 2 + 2x − y 2 + 6y = 15
6. 𝑥 2 − 𝑦 2 + 4 = 0 6. x 2 = y − 1
Circle:
General Form: x2 + y2 + Dx + Ey + F = 0
Standard Form: (x-h)2 + (y-k)2 = r2
Where (h, k) are the coordinates of the center and r is the radius.
The equation of a circle with a center at the origin (0, 0) is x 2 + y2 = r2
Transforming the Equations of a Circle:
I. General form to standard form:
1. Group the terms with the same variable.
2. Move the constant term to the right side of the equation.
3. Create two perfect square trinomials by completing the square method.
Whatever terms that were added to the left side, add them also to the right
side of the equation.
4. Simplify both sides of the equation.
5. Transform the perfect square trinomials into square of binomials to make the
equation in the center-radius form.
Try this:
Transform the General Form to Standard Form.
1. 2x2 + 2y2 + 12x + 20y = 174
2. x2 + y2 – 10x - 12y – 3 = 0
3. 3x2 + 3y2 – 3x – 36y – 9 = 0
II. Standard Form to general form:
1. Square the two binomials.
2. Place all terms to the left side of the equation.
3. Simplify the equation by combining all like terms.
4. Arrange according to the General Form: x2 + y2 + Dx + Ey + F = 0
Try this:
Transform the Standard Form to General From.
1. (y-1)2 + (x+5)2 = 1
2. (x-0)2 + (y-0)2 = 49
3. (x-3)2 + (y-5)2 = 16
Radius = √(𝑥 1 – x2)2 + (y1 – y2)2
In case we are given the center and a point, we will use the distance formula in order to
solve od the radius of the circle.
(x1, y1) are the coordinates of the center,
(x2, y2) are the coordinates of given point.
Try this:
1. Find the equation of a circle in standard form whose center is at (-4, 5) and has a
radius of 13 units.
2. Find the equation of a circle in standard form whose center is at the origin and
passes through the point (4, 3).
Parabola
Opens upward/ downward
Formula:
Standard equation: ( x – h )2 = ± 4𝑝 (𝑦 − 𝑘)
General equation: Ax2 + Cx + Dy + E = 0
(where A & D are nonzero, A > 0)
Vertex: V (h, k)
Axis of Symmetry: x = h
Focus: F (h, k+c)
Line of directrix: ℓ: 𝑦 = 𝑘 − 𝑐
Latus rectum or Focal width: |4c|
Focal length: c
Opens to the left/right
Formula:
Standard equation: ( y – k )2 = 4c ( x – h)
General equation: By2 + Cx + Dy + E = 0
(where B & C are nonzero, B > 0)
Vertex: V ( h, k)
Axis of Symmetry: y = k
Focus: F ( h + c, k)
Line of directrix: ℓ: 𝑥 = ℎ + 𝑐
Latus rectum: |4c|
Focal length: c
Equation Vertex Axis of symmetry Focus Directrix
x2 = 8y
y2 = -28x
(x – 7)2 = 16(y+5)
x = y2 + 4y + 2
5x2 + 30x + 24y = 51
Write the standard equation of the parabola with given conditions.
1. Vertex at the origin and Focus at (5,0).
2. Vertex at the origin and Focus at (0, 4)
3. V (1, -9) and F (-3, -9)
4. 𝑉(5, −4) 𝑎𝑛𝑑 𝐹(3, −4)
5. 𝐹(7,9)𝑎𝑛𝑑 𝑑𝑖𝑟𝑒𝑐𝑡𝑟𝑖𝑥 𝑖𝑠 𝑦 = 3.
6.f (11, 9) and directrix is x = 3.
Find what is asked.
1. A satellite dish has a shape called a paraboloid, where each cross-section is a parabola.
Since radio signals (parallel to the axis) will bounce off the surface of the dish to the focus,
the receiver should be placed at the focus. How far should the receiver be from the vertex,
if the dish is 12 𝑓𝑡 across, and 4.5 𝑓𝑡 deep at the vertex?
2. The cable of a suspension bridge hangs in the shape of a parabola. The towers
supporting the cable are 400 𝑓𝑡 apart and 150 𝑓𝑡 high. If the cable, at its lowest, is 30 𝑓𝑡
above the bridge at its midpoint, how high is the cable 50 𝑓𝑡 away (horizontally) from either
tower?
1. Write the standard form of equation of a parabola whose vertex is at (2, − 3) and
has a directrix at 𝑥 = 5.
2. Find the general form of equation of a parabola whose vertex is at the origin, has a
focal length of 7, has a vertical axis of symmetry, and opens downward.
3. Write the standard form of equation of a parabola whose vertex is at the origin, has a
focal length of 5, has a horizontal axis of symmetry, and opens to the right.
4. The latus rectum of a parabola has its endpoints at (−6, 3)and (6, 3)and its
directrix is 𝑦 = −3.
5. The vertex of a parabola is at (5, 0). It opens upward and has a focal length of 10
units.
6. The vertex of a parabola is at the origin and its focus is at (0, − 4).
Ellipse
Major axis: 2a
Semi-major: a
Minor axis: 2b
Semi-minor: b
Try this:
1. Transform the equation of an ellipse from general form to standard form.
4𝑥 2 + 9𝑦 2 − 36 = 0
2. Find the equation of an ellipse with center at the origin, vertices at (±17, 0)and
covertices at (0, ± 15). Express the answer in standard form.
3. Express the general form of equation of a vertical ellipse with its center at (1, −5),
with 𝑎2 = 3 and 𝑏2 = 2.
4. Write the standard form of equation of an ellipse that has vertices at (5, 3) and
(−5, 3) and has co-vertices at (0, 6) and (0, 0).
5. An ellipse with center at the origin, vertices at (0, ± 4), and a minor axis that is 6
units long.
6. A vertical ellipse with center at the origin, a major axis that is 8 units long, and a
minor axis with endpoints at (±2, 0).
7. An ellipse whose center is at (−1, 2 ), co-vertices at (−1, 4)and (−1, 0) and whose
foci are 5 units away from the center.
Parabola Center Vertices Covertices Foci
𝑥2 𝑦2
+ =1
144 169
(𝑥 + 7)2 (𝑦 − 4)2
+ =1
16 25
9x2 + 16y2 + 72x – 96y + 144 = 0
Find the general equation of the ellipse which satisfies the given conditions.
1. Foci F1 (-7, 6) and F2 (-1, 6), the sum of the distances of any point from the foci is
14.
2. Foci F1 (-3, -2) and F2 (-3, 10), the sum of the distances of any point form the foci is
20.
