Question

Distance between the points on the curve $4x^2+9y^2=1,$ where tangent is parallel to the line $8x=9y,$ is $\frac{2}{\sqrt{k}},$ unit, then $k=$

Answer 1

Given, $4x^2+9y^2=1$
$\Rightarrow \frac{x^2}{\left(\frac{1}{4}\right)}+\frac{y^2}{\left(\frac{1}{9}\right)}=1$
Equation of tangents at $P(\frac{\cos \theta }{2},\frac{\sin \theta }{3})$ is
$2x\cos \theta +3y\sin \theta =1\cdots \left(1\right)$
This is parallel to the given line $8x=9y$
So, slope of line $\left(1\right)$ is $\frac{8}{9}$
$\Rightarrow -\frac{2}{3}\cot \theta =\frac{8}{9}\Rightarrow \tan \theta =-\frac{3}{4}\therefore \cos \theta =\pm \frac{4}{5},\sin \theta =\mp \frac{3}{5}$

Hence, points are $P(\frac{2}{5},-\frac{1}{5})$ and $Q(-\frac{2}{5},\frac{1}{5})$.

Distance between the points $PQ$ is
$\sqrt{\frac{16}{25}+\frac{4}{25}}=\frac{2}{\sqrt{5}}$ unit
Hence, $k=5$