JEE MAIN - Mathematics (2022 - 27th July Morning Shift - No. 22)

If the length of the latus rectum of the ellipse $$x^{2}+4 y^{2}+2 x+8 y-\lambda=0$$ is 4 , and $$l$$ is the length of its major axis, then $$\lambda+l$$ is equal to ____________.
Answer
75

Explanation

Equation of ellipse is : $${x^2} + 4{y^2} + 2x + 8y - \lambda = 0$$

$${(x + 1)^2} + 4{(y + 1)^2} = \lambda + 5$$

$${{{{(x + 1)}^2}} \over {\lambda + 5}} + {{{{(y + 1)}^2}} \over {\left( {{{\lambda + 5} \over 4}} \right)}} = 1$$

Length of latus rectum $$ = {{2\,.\,\left( {{{\lambda + 5} \over 4}} \right)} \over {\sqrt {\lambda + 5} }} = 4$$.

$$\therefore$$ $$\lambda = 59$$.

Length of major axis $$ = 2\,.\,\sqrt {\lambda + 5} = 16 = l$$

$$\therefore$$ $$\lambda + l = 75$$.

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