Conic Sections Worksheet Answer Key – Precalculus

Problem 1 Circle from Its Equation

Compare with (x − h)² + (y − k)² = r². Here h = 6 and k = −1, and r² = 20, so r = √20 = 2√5 ≈ 4.47.

Answer: center (6, −1), radius 2√5

Problem 2 Circle from a Point

r² = (−1 − (−7))² + (−1 − 4)² = 36 + 25 = 61.

Answer: (x + 7)² + (y − 4)² = 61

Problem 3 Focus and Directrix

Compare with x² = 4py: 4p = 12, so p = 3. The vertex is (0, 0), the focus is (0, 3) and the directrix is y = −3.

Answer: vertex (0, 0), focus (0, 3), directrix y = −3

Problem 4 A Sideways Parabola

4p = −20 gives p = −5. The focus is (−5, 0) and the directrix is x = 5. Since p is negative, the parabola opens to the left.

Answer: focus (−5, 0), directrix x = 5, opens left

Problem 5 Equation from the Figure

The focus is 4 units below the vertex, so p = −4. The equation is x² = 4py = −16y. The directrix is y = 4.

Answer: x² = −16y

Problem 6 Parts of an Ellipse

a² = 49 (under x), so a = 7, and b² = 25, so b = 5. The major axis is horizontal.

c² = 49 − 25 = 24, so c = 2√6 ≈ 4.90.

Answer: vertices (±7, 0), minor axis endpoints (0, ±5), foci (±2√6, 0)

Problem 7 Ellipse from Vertices and Foci

The major axis is vertical with a = 9. The focal distance is c = 5, so b² = a² − c² = 81 − 25 = 56.

Answer: x²56 + y²81 = 1

Problem 8 Parts of a Hyperbola

a = 5 and b = √11. c² = 25 + 11 = 36, so c = 6. The main axis is horizontal.

The asymptotes are y = ±√115x.

Answer: vertices (±5, 0), foci (±6, 0), asymptotes y = ±√115x

Problem 9 A Vertical Hyperbola

a² = 9 so a = 3, and b² = 27 so b = 3√3. c² = 9 + 27 = 36, so c = 6. The main axis is vertical.

The asymptotes are y = ±abx = ±33√3x = ±√33x.

Answer: vertices (0, ±3), foci (0, ±6), asymptotes y = ±√33x

Problem 10 Shifted Parabola

The focus is 5 units above the vertex, so p = 5. The equation is (x − 5)² = 4(5)(y + 2) = 20(y + 2). The directrix is y = −2 − 5 = −7.

Answer: (x − 5)² = 20(y + 2), directrix y = −7

Problem 11 Completing the Square: Circle

Group: (x² − 8x) + (y² + 10y) = −5. Add 16 and 25 to both sides: (x − 4)² + (y + 5)² = −5 + 16 + 25 = 36.

Answer: (x − 4)² + (y + 5)² = 36, center (4, −5), radius 6

Problem 12 Completing the Square: Parabola

Move the x term and constant: y² − 4y = 12x − 40. Add 4: (y − 2)² = 12x − 36 = 12(x − 3).

So 4p = 12 and p = 3. The vertex is (3, 2), the focus is (6, 2) and the directrix is x = 0.

Answer: (y − 2)² = 12(x − 3), vertex (3, 2), focus (6, 2), directrix x = 0

Problem 13 Completing the Square: Ellipse

9(x² − 2x) + 4(y² + 4y) = 11. Complete the squares: 9(x − 1)² + 4(y + 2)² = 11 + 9 + 16 = 36.

Divide by 36: (x − 1)²4 + (y + 2)²9 = 1. It is an ellipse with center (1, −2), vertical major axis, a = 3, b = 2 and c = √5.

Answer: ellipse; center (1, −2); vertices (1, 1) and (1, −5); foci (1, −2 ± √5)

Problem 14 Completing the Square: Hyperbola

4(x² + 6x) − 5(y² − 2y) = −11. Complete the squares, remembering the factors: 4(x + 3)² − 5(y − 1)² = −11 + 36 − 5 = 20.

Divide by 20: (x + 3)²5 − (y − 1)²4 = 1. So a² = 5, b² = 4 and c² = 9, c = 3. The center is (−3, 1) with a horizontal main axis.

Foci: (−3 ± 3, 1) = (0, 1) and (−6, 1). Asymptotes: y − 1 = ±2√5(x + 3).

Answer: center (−3, 1), foci (0, 1) and (−6, 1), asymptotes y − 1 = ±2√5(x + 3)

Problem 15 Satellite Dish

(a) Use x² = 4py. The rim point is (5, 1.25), so 25 = 4p(1.25) = 5p and p = 5. The equation is x² = 20y.

(b) The focus is at (0, p) = (0, 5), so the receiver sits 5 feet above the vertex.

Answer: (a) x² = 20y; (b) 5 feet above the vertex

Problem 16 Whispering Hall

(a) a = 40 and b = 15, so x²1600 + y²225 = 1.

(b) c² = 1600 − 225 = 1375, so c = √1375 ≈ 37.1 feet.

(c) The foci are 2c ≈ 74.2 feet apart.

Answer: (a) x²1600 + y²225 = 1; (b) about 37.1 ft; (c) about 74.2 ft

Problem 17 Elliptical Bridge Arch

(a) a = 30 and b = 18: x²900 + y²324 = 1 with y ≥ 0.

(b) At x = 20: y = 18√(1 − 400900) = 18√59 = 6√5 ≈ 13.4 feet.

(c) At y = 14: x = 30√(1 − 196324) = 30√128324 ≈ 18.9. The truck can be at most 2(18.9) ≈ 37.7 feet wide.

Answer: (a) x²900 + y²324 = 1; (b) about 13.4 ft; (c) about 37.7 ft

Problem 18 Radio Station Hyperbola

(a) The foci are at (±90, 0), so c = 90. The difference of distances is 2a = 100, so a = 50. Then b² = c² − a² = 8100 − 2500 = 5600. The equation is x²2500 − y²5600 = 1.

(b) At the vertices (±50, 0).

(c) y = ±√560050x = ±20√1450x = ±2√145x ≈ ±1.50x.

Answer: (a) x²2500 − y²5600 = 1; (b) (±50, 0); (c) y ≈ ±1.50x

Problem 19 A Comet’s Orbit

(a) The major axis is 2a = 52, so a = 26. The closest distance is a − c = 10, so c = 16.

(b) b² = a² − c² = 676 − 256 = 420. The equation is x²676 + y²420 = 1.

(c) The greatest distance is a + c = 42 million miles.

Answer: (a) a = 26, c = 16; (b) x²676 + y²420 = 1; (c) 42 million miles

Problem 20 Identify and Describe

Group: 4(x² − 6x) − 9(y² + 4y) = 36. Complete the squares: 4(x − 3)² − 9(y + 2)² = 36 + 36 − 36 = 36.

Divide by 36: (x − 3)²9 − (y + 2)²4 = 1. The minus sign makes it a hyperbola with center (3, −2), a = 3, b = 2 and c = √13.

Vertices: (0, −2) and (6, −2). Foci: (3 ± √13, −2). Asymptotes: y + 2 = ±23(x − 3).

Answer: (x − 3)²9 − (y + 2)²4 = 1; center (3, −2); vertices (0, −2), (6, −2); foci (3 ± √13, −2)