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

Answer 1

Answer: (B), (D)

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

We have ${4x}^2+9y^2=1$ ...(1)
Differentiating w.r.t x, we get
$8x+18y\frac{dy}{dx}=0\Rightarrow \frac{dy}{dx}=-\frac{4x}{9y}$
The tangent at point $(x,y)$ will be parallel to the line $8x=9y$ if
$\frac{-4x}{9y}=\frac{8}{9}\Rightarrow x=-2y$
Substituting this in (1), we get
$4{(-2y)}^2+{9y}^2=1\Rightarrow {25y}^2=1\Rightarrow y=\pm \frac{1}{5}$
Thus the point where the tangent are parallel to $8x=9y$ are
$(\frac{-2}{5},\frac{1}{5})$ and $(\frac{2}{5},\frac{-1}{5})$