Question

If the radii of director circles of `x^2/a^2 + y^2/b^2` = 1 and `x^2/a^2 - y^2/b^2` = (a > b) are 2r and r respectively, then `e_2^2/e_1^2` is equal to ______.

(where e₁, e₂ are their eccentricities respectively)

Options

• 1

• 2

• 3

• 4

Answer 1

If the radii of director circles of `x^2/a^2 + y^2/b^2` = 1 and `x^2/a^2 - y^2/b^2` = (a > b) are 2r and r respectively, then `e_2^2/e_1^2` is equal to 4.

(where e₁, e₂ are their eccentricities respectively)

Explanation:

a² + b² = r²

a² − b² = `r^2/4`

a² = `(5r^2)/8` and b² = `(3r^2)/8`

b² = `a^2(1 - e_1^2)`, b² = `a^2(e_2^2 - 1)`

⇒ `e_1^2 = 2/5` and `e_2^2 = 8/5`

Now, `e_2^2/e_1^2 = 8/2` = 4