Question

If the tangents on the ellipse $4x^{2}+y^{2}=8$ at the points $\left(1 , 2\right)$ and $\left(a , b\right)$ are perpendicular to each other, then $a^{2}$ is equal to :

• A. $\frac{64}{17}$
• B. $\frac{2}{17}$
• C. $\frac{128}{17}$
• D. $\frac{4}{17}$

Answer 1

Answer: (B)

$4a^{2}+b^{2}=8...\left(1\right)$ , because point $\left(a , b\right)$ lies on ellipse.
Also $\left(\frac{d y}{d x}\right)_{\left(1 , 2\right)}=-\frac{4 x}{y}=-2$
$\Rightarrow -\frac{4 a}{b}=\frac{1}{2}$ , tangents at $\left(1 , 2\right)\&\left(a , b\right)$ are perpendicular to each other.
$b=-8a$
$\Rightarrow b^{2}=64a^{2}$
$68a^{2}=8$
$a^{2}=\frac{2}{17}$