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10-2. Parabola, Ellipse, Hyperbola
hard
If the tangents on the ellipse $4x^2 + y^2 = 8$ at the points $(1, 2)$ and $(a, b)$ are perpendicular to each other, then $a^2$ is equal to
A
$\frac{2}{{17}}$
B
$\frac{4}{{17}}$
C
$\frac{64}{{17}}$
D
$\frac{128}{{17}}$
(JEE MAIN-2019)
Solution
$4{a^2} + {b^2} = 8\,\,\,\,\,\,…….\left( 1 \right)$
${\left( {\frac{{dy}}{{dx}}} \right)_{\left( {1,2} \right)}} = – \frac{{4x}}{y} = – 2$
$ \Rightarrow – \frac{{4a}}{b} = \frac{1}{2}$
$b = – 8a$
$ \Rightarrow {b^2} = 64{a^2}$
$68{a^2} = 8,{a^2} = \frac{2}{{17}}$
Standard 11
Mathematics
