Question

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

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

Answer 1

The correct option is A $\frac{2}{17}$

Given, equation of ellipse
$4x^2+y^2=8$
$\frac{dy}{dx}=-\frac{4x}{y}$

Since, tangent at $(1,2)$ and $(a,b)$ are perpendicular.
$\therefore (-\frac{4}{2})(-\frac{4a}{b})=-1$
$\Rightarrow b=-8a...\left(1\right)$

$(a,b)$ lies on ellipse
$\Rightarrow 4a^2+b^2=8$
$\Rightarrow 4a^2+64a^2=8$
$\Rightarrow a^2=\frac{8}{68}=\frac{2}{17}$