Question

Find the equations of the tangent and the normal to the following curves.
$x^2+y^2+xy=3atP(1,1)$

Answer 1

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

Differentiating w.r.t. $x$ , we get

$2x+2y\frac{dy}{dx}+x\frac{dy}{dx}+y=0$

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

$\frac{dy}{dx}=\frac{-2x-y}{2y+x}$

Slope of the tangent at $(1,1)$ is $m={\left(\frac{dy}{dx}\right)}_{(1,1)}=\frac{-3}{3}=-1$

So, the equation of the tangent is

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

$y-1=-x+1$

$y+x=2$

Slope of the normal at $(1,1)$ is $=-\frac{1}{m}=1$

So, the equation of the normal is

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

$y=x$