Chapter 10: Conic Sections

Foci (0, ± 13), the conjugate axis is of length 24

Solutions:

The foci lie on the y-axis, so the equation is of the form y²/a² − x²/b² = 1.

Length of the conjugate axis = 2b = 24, so b = 12. Also c = 13.

Then a² = c² − b² = 169 − 144 = 25.

Required equation: y²/25 − x²/144 = 1

Foci (± 3√5, 0), the latus rectum is of length 8

Solutions:

The foci lie on the x-axis, so the equation is of the form x²/a² − y²/b² = 1, with c = 3√5.

Length of the latus rectum = 2b²/a = 8, which gives b² = 4a.

Using c² = a² + b²: 45 = a² + 4a

a² + 4a − 45 = 0 ⟹ (a + 9)(a − 5) = 0 ⟹ a = 5 (rejecting a = −9, as a cannot be negative)

Then b² = 4 × 5 = 20.

Required equation: x²/25 − y²/20 = 1

Foci (± 4, 0), the latus rectum is of length 12

Solutions:

The foci lie on the x-axis, so the equation is of the form x²/a² − y²/b² = 1, with c = 4.

Latus rectum: 2b²/a = 12, so b² = 6a.

Using c² = a² + b²: 16 = a² + 6a

a² + 6a − 16 = 0 ⟹ (a + 8)(a − 2) = 0 ⟹ a = 2 (rejecting a = −8)

Then b² = 6 × 2 = 12.

Required equation: x²/4 − y²/12 = 1

Vertices (± 7, 0), e = 4/3

Solutions:

The vertices lie on the x-axis, so the equation is of the form x²/a² − y²/b² = 1, with a = 7.

Since e = c/a = 4/3, we get c = 7 × 4/3 = 28/3.

Then b² = c² − a² = 784/9 − 49 = (784 − 441)/9 = 343/9.

Required equation: x²/49 − y²/(343/9) = 1, i.e., x²/49 − 9y²/343 = 1

Foci (0, ± √10), passing through (2, 3)

Solutions:

The foci lie on the y-axis, so the equation is of the form y²/a² − x²/b² = 1, with c = √10, i.e., c² = 10, so b² = 10 − a².

Since the hyperbola passes through (2, 3):

9/a² − 4/(10 − a²) = 1

9(10 − a²) − 4a² = a²(10 − a²)

90 − 9a² − 4a² = 10a² − a⁴

a⁴ − 23a² + 90 = 0

Solving as a quadratic in a²: a² = [23 ± √(529 − 360)]/2 = (23 ± 13)/2, giving a² = 18 or a² = 5.

Since a² must be less than c² = 10, we take a² = 5, and then b² = 10 − 5 = 5.

Required equation: y²/5 − x²/5 = 1, i.e., y² − x² = 5