Medium INTEGER +4 / -1 PYQ · JEE Mains 2022

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 (integer) 75

Solution

<p>Equation of ellipse is : ${x^2} + 4{y^2} + 2x + 8y - \lambda = 0$</p> <p>${(x + 1)^2} + 4{(y + 1)^2} = \lambda + 5$</p> <p>$${{{{(x + 1)}^2}} \over {\lambda + 5}} + {{{{(y + 1)}^2}} \over {\left( {{{\lambda + 5} \over 4}} \right)}} = 1$$</p> <p>Length of latus rectum $$ = {{2\,.\,\left( {{{\lambda + 5} \over 4}} \right)} \over {\sqrt {\lambda + 5} }} = 4$$.</p> <p>$\therefore$ $\lambda = 59$.</p> <p>Length of major axis $= 2\,.\,\sqrt {\lambda + 5} = 16 = l$</p> <p>$\therefore$ $\lambda + l = 75$.</p>

About this question

Subject: Mathematics · Chapter: Conic Sections · Topic: Parabola

This question is part of PrepWiser's free JEE Main question bank. 146 more solved questions on Conic Sections are available — start with the harder ones if your accuracy is >70%.

Drill 25 more like these. Every day. Free.

PrepWiser turns these solved questions into a daily practice loop. Chapter-wise drills, full mocks, AI doubt chat. No auto-renew.

Start free →