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. none of these

Answer 1

Answer: (B)

We have $P{F}_1+P{F}_2=2a=10$
for every point $P$ on the ellipse.
Differentiating w.r.t. $x$, we get
$8x+18y\frac{dy}{dx}=0$
$\Rightarrow \frac{dy}{dx}=-\frac{8x}{18y}=-\frac{4x}{9y}$
The tangent at point $(x,y)$ will be parallel to $8x=9y$ if
$\frac{-4x}{9y}=\frac{8}{9}\Rightarrow x=-2y$
Substituting $x=-2y$ in $4x^2+9y^2=1$, we get
$4(-2y{)}^2+9y^2=1\text{ or }25y^2=1$
$\Rightarrow y=\pm \frac{1}{5}$
Thus, the points where the tangents are parallel to
$8x=$ $9y$ are $\left(-\frac{2}{5},\frac{1}{5}\right)$ and $\left(\frac{2}{5},-\frac{1}{5}\right)$