Question

Find the equation of the tangents to the ellipse $4x^2+9y^2=72$ which is perpendicular to the line $3x-2y=5$

Answer 1

We will first find the slope of the tangent to the ellipse $4x^2+9y^2=72$ using differentiation, as follows:

$4x^2+9y^2=72$

$8x+18y\left(\frac{dy}{dx}\right)=0$

$\left(\frac{dy}{dx}\right)=\frac{-4x}{9y}$

Now the slope of the given line $3x-2y=5$ is

$\Rightarrow y=\left(\frac{3}{2}\right)x-\frac{5}{2}$

$\Rightarrow$ slope of this line is $\frac{3}{2}$

Now, this slope is perpendicular to the tangent of the ellipse, so we have:

$\frac{-4x}{9y}\ast \frac{3}{2}=-1$

$\to \frac{-4x}{9y}=\frac{-2}{3}$

$\to x=\frac{3}{2}y$

Now , we will put this value of $x$ in the equation of the ellipse to get the point at which the tangent is to be found.

$4x^2+9y^2=72$

$4(\frac{3}{2}y{)}^2+9y^2=72$

$18y^2=72$

$\Rightarrow y=\pm 2$

$\Rightarrow x=\frac{3}{2}y=\pm 3$

We also have the slope of the tangent to ellipse as $\frac{-2}{3}$

Thus the equations of the tangents at $(3,2)$ is :

$y-2=\frac{-2}{3}(x-3)$

$\Rightarrow 3y=-2x+12$

and another equation of the tangent at $(-3,-2)$ is :

$y+2=\frac{-2}{3}(x+3)$

$\Rightarrow 3y=-2x-12$