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Q.
On the ellipse $4 x^{2}+9 y^{2}=1$, the points at which the tangents are parallel to the line $8 x=9 y$ are:
Conic Sections
Solution:
We have $P F_{1}+P F_{2}=2 a=10$
for every point $P$ on the ellipse.
Differentiating w.r.t. $x$, we get
$8 x+18 y \frac{d y}{d x}=0 $
$\Rightarrow \frac{d y}{d x}=-\frac{8 x}{18 y}=-\frac{4 x}{9 y}$
The tangent at point $(x, y)$ will be parallel to $8 x=9 y$ if
$\frac{-4 x}{9 y}=\frac{8}{9} \Rightarrow x=-2 y$
Substituting $x=-2 y$ in $4 x^{2}+9 y^{2}=1$, we get
$4(-2 y)^{2}+9 y^{2}=1 \text { or } 25 y^{2}=1$
$ \Rightarrow y=\pm \frac{1}{5}$
Thus, the points where the tangents are parallel to
$8 x=$ $9 y$ are $\left(-\frac{2}{5}, \frac{1}{5}\right)$ and $\left(\frac{2}{5},-\frac{1}{5}\right)$