Question

Let the eccentricity of an ellipse `x^2/a^2 + y^2/b^2` = 1, a > b, be `1/4`. If this ellipse passes through the point ```(-4sqrt(2/5), 3)`, then a² + b² is equal to ______.

Options

• 29

• 31

• 32

• 34

Answer 1

Let the eccentricity of an ellipse `x^2/"a"^2 + "y"^2/"b"^2` = 1, a > b, be `1/4`. If this ellipse passes through the point ```(-4sqrt(2/5), 3)`, then a² + b² is equal to 31.

Explanation:

Equation of ellipse is `x^2/"a"^2 + "y"^2/"b"^2` = 1, a > b.

Given eccentricity is `1/4`.

⇒ e² = `1 - "b"^2/"a"^2`

`1/16 = 1- "b"^2/"a"^2`

⇒ `"b"^2/"a"^2` = `1 - 1/16` = `15/16`

⇒ b² = `15/16"a"^2`

Put the value of b² in the equation of ellipse and satisfy the point `(-4sqrt(2/5), 3)`.

⇒ `(16 xx 2/5)/"a"^2 + 9/"b"^2` = 1

⇒ `32/(5a^2) + 9/b^2` = 1

⇒ `32/(5"a"^2) + 9/(15/16"a"^2)` = 1

⇒ `80/(5"a"^2)` = 1

⇒ 16 = a²

Put the value of a² in b².

⇒ b² = `15/16"a"^2`

b² = 15

Add a² and b²; a² + b² = 16 + 15 = 31.