Question

The normal to the curve $x^2+2xy-3y^2=0$ at $(1,1)$

Answer 1

Given the curve

$x^2+2xy-3y^2=0$

On differentiating this equation with respect to x and we get,

$2x+2(x\frac{dy}{d}+y)-3\times 2y\frac{dy}{dx}=0$

$\Rightarrow x+x\frac{dy}{dx}+y-3y\frac{dy}{dx}=0$

$\Rightarrow \frac{dy}{dx}(x-3y)=-(x+y)$

$\Rightarrow \frac{dy}{dx}=\frac{(x+y)}{(3y-x)}$

Slope at point (1,1)

$\Rightarrow \frac{dy}{dx}={\left[\frac{(x+y)}{(3y-x)}\right]}_{(1,1)}$

$\Rightarrow \frac{dy}{dx}={\left[\frac{(1+1)}{(3\times 1-1)}\right]}_{(1,1)}$

$\Rightarrow \frac{dy}{dx}=\left[\frac{2}{2}\right]$

$\Rightarrow \frac{dy}{dx}=1$

Slope of normal

$-\frac{dx}{dy}=\frac{-1}{\frac{dy}{dx}}=\frac{-1}{1}=-1$

$-\frac{dx}{dy}=-1$

We know that equation of normal

$y-y_1=-\frac{dx}{dy}(x-x_1)$

At point $(1,1)$

$y-1=-1(x-1)$

$\Rightarrow y-1=-x+1$

$\Rightarrow x+y-2=0$

Hence , this is the answer.