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)

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