Q. On the ellipse $2x^{2} + 3y^{2} = 1$ the points at which the tangent is parallel to $4x = 3y + 4,$ are
A) $\left( \frac{2}{\sqrt{11}},\frac{1}{\sqrt{11}} \right)$ or $\left( - \frac{2}{\sqrt{11}}, - \frac{1}{\sqrt{11}} \right)$
B) $\left( - \frac{2}{\sqrt{11}},\frac{1}{\sqrt{11}} \right)$ or $\left( \frac{2}{\sqrt{11}}, - \frac{1}{\sqrt{11}} \right)$
C) $\left( - \frac{2}{5},\ - \frac{1}{5} \right)$
D) $\left( - \frac{3}{5},\ - \frac{2}{5} \right)$ or $\left( \frac{3}{5},\ \frac{2}{5} \right)$
Explanation
(b)
We have, $a^{2} = \frac{1}{2},\ b^{2} = \frac{1}{3},\ m = \frac{4}{3}$
The required points are
$\left( \pm \frac{a^{2}m}{\sqrt{a^{2}m^{2} + b^{2}}},\ \pm \frac{b^{2}}{\sqrt{a^{2}m^{2} + b^{2}}} \right)$
$= \left( \pm \frac{\frac{1}{2} \times \frac{4}{3}}{\sqrt{\frac{1}{2} \times \frac{16}{9} + \frac{1}{3}}}, \pm \frac{\frac{1}{3}}{\sqrt{\frac{1}{2} \times \frac{16}{9} + \frac{1}{3}}} \right) = \left( \pm \frac{2}{\sqrt{11}},\ \pm \frac{1}{\sqrt{11}} \right)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.