Question

Find the equation to the tangent and normal at the point $(1,\frac{4}{3})$ of the ellipse $4x^2+9y^2=20$.

Answer 1

Given equation of ellipse is $4x^2+9y^2=20$ ...... $\left(i\right)$

Slope of the tangent is $\frac{dy}{dx}$ at that point,

From $\left(i\right)$, we get

$8x+18y\frac{dy}{dx}=0$

$\Rightarrow \frac{dy}{dx}=-\frac{4x}{9y}=-\frac{4\times 3}{9\times 4}=-\frac{1}{3}$

Equation of tangent,

$y-\frac{4}{3}=-\frac{1}{3}(x-1)$

$x+3y=5$

Slope of normal will be $\frac{-1}{\frac{dy}{dx}}=\frac{-1}{\frac{-1}{3}}=3$

Equation of normal will be,

$y-\frac{4}{3}=3(x-1)$

$y-\frac{4}{3}=3x-3$

$9x-3y=5$