Question

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

Answer 1

$\begin{array}{c}4x^2+9y^2=1\\ 8xdx+8ydy=0\left(differentiate\right)\\ \frac{dy}{dx}=-\left(\frac{4x}{9y}\right)\end{array}$

the linear equation giving the line tangent to the ellipse at any given point $(x,y)$

$y=-\left(\frac{4x}{9y}\right)x+b$

the slopes of tnagent line and the above equation must be same

$\begin{array}{c}-\left(\frac{4x}{9y}\right)=\frac{8}{9}\\ \left(-0.5x\right)=y\end{array}$

now substitute this into $y=\frac{8}{9}x-\frac{5}{9}$

$\begin{array}{c}\left(-0.5x\right)=\left(\frac{8}{9}\right)x-\frac{5}{9}\\ \left(\frac{25}{18}\right)x=\frac{5}{9}\\ x=\frac{2}{5}\end{array}$

using the value of $x$

$\begin{array}{c}\left(-0.5\times \frac{2}{5}\right)=y\\ y=-\frac{1}{5}\end{array}$

hence $\left(\frac{2}{5},\frac{-1}{5}\right)$ this is the point where the line $y=\frac{8}{9}x-\frac{5}{9}$ is the tangent to the specified ellipse