Question

On the ellipse $4x^2+9y^2=1$, the points at which the tangents are parallel to the line $8x=9y$ are

• A. $\left(\frac{2}{5},\frac{1}{5}\right)$
• B. $\left(-\frac{2}{5},\frac{1}{5}\right)$
• C. $\left(-\frac{2}{5},-\frac{1}{5}\right)$
• D. $\left(\frac{2}{5},-\frac{1}{5}\right)$

Answer 1

Answer: (B), (D)

Let $\left(\frac{1}{2}cos\theta ,\frac{1}{3}sin\theta \right)$ be a point on $4x^2+9y^2=1$, so equation of tangent at
$2xcos\theta +3ysin\theta =1$
equating slope with $8x=9y$
$\frac{-2cos\theta }{3sin\theta }=\frac{8}{9}\Rightarrow tan\theta =-\frac{3}{4}$
Hence either $cos\theta =-\frac{4}{5},sin\theta =-\frac{3}{5}$
so the points are $\left(-\frac{2}{5},\frac{1}{5}\right)$or $\left(\frac{2}{5},-\frac{1}{5}\right)$