Question

On the ellipse $4x^2+9y^2=1$, the point at which the tangent is parallel to the line $8x=9y$ is

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

Answer 1

Answer: (B)

$(-\frac{2}{5},\frac{1}{5})$

$\because \frac{dy}{dx}=\frac{-4x}{9y}=\frac{8}{9}\&x=-2y$
Substituting $x=-2y$ in $4x^2+9y^2=1$, we get $y=\pm \frac{1}{5}$
Therefore, $(x,y)=(-\frac{2}{5},\frac{1}{5})$
Ans: B