Question

On the ellipse $4x^2+9y^2=1$, the points at which the tangents are parallel to the line $8x=9y$ are

• A. $(\frac{2}{5},\frac{1}{5})$
• B. $(-\frac{2}{5},\frac{1}{5})$
• C. $(-\frac{2}{5},-\frac{1}{5})$
• D. $(\frac{2}{5},-\frac{1}{5})$

Answer 1

Answer: (B), (D)

The correct options are
B $(-\frac{2}{5},\frac{1}{5})$
D $(\frac{2}{5},-\frac{1}{5})$

Any point on the ellipse $4x^2+9y^2=1$
$(\frac{1}{2}\cos \theta ,\frac{1}{3}\sin \theta )$
Slope of the tangent will be,
$8x+18y\frac{dy}{dx}=0\Rightarrow \frac{dy}{dx}=-\frac{2}{3}\frac{\cos \theta }{\sin \theta }$
The tangent is parallel to $8x=9y$
$\frac{8}{9}=-\frac{2}{3}\frac{\cos \theta }{\sin \theta }\Rightarrow \tan \theta =-\frac{4}{3}$
So,
$\Rightarrow \cos \theta =-\frac{4}{5},\sin \theta =\frac{3}{5}$
or
$\Rightarrow \cos \theta =\frac{4}{5},\sin \theta =-\frac{3}{5}$

Corresponding points will be,
$(-\frac{2}{5},\frac{1}{5})$ and $(\frac{2}{5},-\frac{1}{5})$