MCQ
The value of $\lambda $, for which the line $2x - \frac{8}{3}\lambda y = - 3$ is a normal to the conic ${x^2} + \frac{{{y^2}}}{4} = 1$ is
- A$\frac{{\sqrt 3 }}{2}$
- B$\frac{1}{2}$
- C$ - \frac{{\sqrt 3 }}{2}$
- ✓$\frac{3}{8}$
$ax\sin \theta - by{\rm{cosec }}\theta = {a^2} - {b^2}$.....$(i)$
Comparing equation $(i)$ with $2x - \frac{8}{3}\lambda y = - 3$. We get,
$a\sin \theta = 2$, $b{\rm{ cosec}}\theta = \frac{8}{3}\lambda $ or $ab = \frac{{16}}{3}\lambda $.....$(ii)$
$\because \,a = 1,\,b = 2$; $2 = \frac{{16}}{3}\lambda $ or $\lambda = 3/8$
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