Search any question & find its solution
Question:
Answered & Verified by Expert
The value of $\lambda$, for which the line $2 x-\frac{8}{3} \lambda y=-3$ is a normal to the conic $x^2+\frac{y^2}{4}=1$ is
Options:
Solution:
2125 Upvotes
Verified Answer
The correct answer is:
$\frac{\sqrt{3}}{2}$
Equation of normal to ellipse $x^2+\frac{y^2}{4}=1$ is $\frac{x}{\cos \theta}-\frac{2 y}{\sin \theta}=1-4$ (i)
Comparing with $2 x-\frac{8}{3} \lambda y=-3$
$\cos \theta=\frac{1}{2}, \lambda=\frac{3}{4 \sin \theta}=\frac{\sqrt{3}}{2}$
Comparing with $2 x-\frac{8}{3} \lambda y=-3$
$\cos \theta=\frac{1}{2}, \lambda=\frac{3}{4 \sin \theta}=\frac{\sqrt{3}}{2}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.