Question

Find the equation of an ellipse whose foci are at(± 3, 0) and which passes through (4, 1).

Answer 1

Given that we need to find the equation of the ellipse whose foci are at (±3, 0) and passes through (4, 1).

Let us assume the equation of the ellipse is

We know that foci are (±ae, 0) and eccentricity of the ellipse is

⇒ ae = 3

⇒ a² - b² = 9 ..... - - - (2)

Substituting the point (4,1) in (1) we get,

⇒ 16b² + a² = a²b² From (2),

⇒ 16(a² - 9) + a² = a²(a² - 9)

⇒ 16a² - 144 + a² = a⁴ - 9a²

⇒ a⁴ - 26a² + 144 = 0

⇒ a⁴ - 18a² - 8a² + 144 = 0

⇒ a²(a² - 18) - 8(a² - 18) = 0

⇒ (a² - 8)(a² - 18) = 0

⇒ a² - 8 = 0 (or) a² - 18 = 0

⇒ a² = 8 (or) a² = 18

⇒ b² = 18 - 9(since b² > 0)

⇒ b² = 9.

The equation of the ellipse is

⇒ x² + 2y² = 18

∴ The equation of the ellipse is x² + 2y² = 18.