JEE Main 2019 — Ellipse Question with Solution

Focus (0, be) = (0, 5$$\sqrt{3}$$)

$$\therefore$$ be = 5$$\sqrt{3}$$

$$\Rightarrow$$ b²e² = 75

As here b > a

so e² = $$1-\frac{a^2}{b^2}$$

$$\therefore$$ b²$$\left(1-\frac{a^2}{b^2}\right)$$ = 75

$$\Rightarrow$$ b² - a² = 75 .......(1)

Given that,

difference of the lengths of major axis and minor axis is 10.

$$\therefore$$ 2b - 2a = 10

$$\Rightarrow$$ b - a = 5 .......(2)

From (1),

(b + a)(b - a) = 75

$$\Rightarrow$$ (b + a)5 = 75

$$\Rightarrow$$ (b + a) = 15 .......(3)

From (2) and (3) we get,

b = 10, a = 5

$$\therefore$$ Length of its latus rectum = $$\frac{2a^2}{b}$$

= $$\frac{2\times 25}{10}$$ = 5