Mathfolis

Conic Sections

Unit 4 · Functions Involving Parameters, Vectors, and Matrices

What AP Precalc asks here

Four families of conic sections show up in standard form: circles, ellipses, parabolas, and hyperbolas. Recognizing them from an implicit equation is mostly pattern matching: both squared terms with the same sign → ellipse (or circle if coefficients match); both squared but opposite signs → hyperbola; only one squared term → parabola. Completing the square on a general quadratic reveals the center and standard form.

Standard forms

Circle
(xh)2+(yk)2=r2(x - h)^2 + (y - k)^2 = r^2
Ellipse
(xh)2a2+(yk)2b2=1\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1
Parabola (vertical axis)
yk=a(xh)2y - k = a(x - h)^2
Hyperbola
(xh)2a2(yk)2b2=1\frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1
AP Tip: If the equation has both x2x^2 and y2y^2 with the SAME coefficient and the same sign, it's a circle (or empty set). Different coefficients but same sign → ellipse. Opposite signs → hyperbola.
Type 1

Identify the conic from its equation

Look at the squared terms: matching coefficients and same sign → circle; different but same sign → ellipse; opposite signs → hyperbola; one squared → parabola.

Example 1
Identify the conic x216y29=1\dfrac{x^2}{16} - \dfrac{y^2}{9} = 1. (A) Circle (B) Ellipse (C) Parabola (D) Hyperbola

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Type 2

Read center, vertices, axes

h,kh, k from the shifts; semi-axes from the denominators (square root for ellipse/hyperbola).

Example 2
What is the center of the ellipse (x2)29+(y+1)24=1\dfrac{(x - 2)^2}{9} + \dfrac{(y + 1)^2}{4} = 1? (A) (2,1)(-2, 1) (B) (2,1)(2, -1) (C) (2,1)(2, 1) (D) (9,4)(9, 4)

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Type 3

Convert to standard form

Complete the square on the xx terms and on the yy terms.

Example 3
Convert x2+y26x+4y12=0x^2 + y^2 - 6x + 4y - 12 = 0 to standard form. (A) (x3)2+(y+2)2=25(x - 3)^2 + (y + 2)^2 = 25 (B) (x+3)2+(y2)2=25(x + 3)^2 + (y - 2)^2 = 25 (C) (x3)2+(y+2)2=1(x - 3)^2 + (y + 2)^2 = 1 (D) (x3)2+(y+2)2=12(x - 3)^2 + (y + 2)^2 = 12

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