Q. On the ellipse $4x^{2} + 9y^{2} = 1\$the point at which the tangent are parallel to$\ 8x = 9y$ are
A) $\left( \frac{2}{5},\frac{1}{5} \right)$ or $\left( - \frac{2}{5},\frac{1}{5} \right)$
B) $\left( - \frac{2}{5},\frac{1}{5} \right)$ or $\left( \frac{2}{5}, - \frac{1}{5} \right)$
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)
Here, $a^{2} = - \frac{1}{4},\ b^{2}, = \frac{1}{9},\ m = \frac{8}{9}$
$\therefore Point\ of\ contact\ is\$
$\left( \pm \frac{a^{2}m}{\sqrt{a^{2}m^{2} + b^{2}}}, \mp \frac{b^{2}}{\sqrt{a^{2}m^{2} + b^{2}}} \right)$
$= \left( \pm \frac{\frac{1}{4}.\frac{8}{9}}{\sqrt{\frac{1}{4} \times \frac{64}{81} + \frac{1}{9}}},\ \pm \frac{\frac{1}{9}}{\sqrt{\frac{1}{4} \times \frac{64}{81} + \frac{1}{9}}} \right)$
$= \left( \pm \frac{2}{5}, \pm \frac{1}{5} \right)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.