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
$2x\, cos\,\theta +3\,y\,sin\,\theta =1$
equating slope with $8x = 9y$
$\frac{-2\,cos\,\theta}{3\,sin\,\theta}=\frac{8}{9} \Rightarrow tan\,\theta=-\frac{3}{4}$
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 $\left(\frac{2}{5}, -\frac{1}{5}\right)$