Question

Find the slopes of the tangent and the normal to the following curves at the indicated points.
$x^2+3y+y^2=5$ at $(1,1)$.

Answer 1

We have,

$x^2+3y+y^2=5$

Now, differentiate w.r.t $x$, we get

$2x+3\frac{dy}{dx}+2y\frac{dy}{dx}=0$

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

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

$\frac{dy}{dx}=-\frac{2x}{3+2y}$

Put $x=1,y=1$

So,

${\frac{dy}{dx}}_{(1,1)}=-\frac{2\left(1\right)}{3+2\left(1\right)}$

${\frac{dy}{dx}}_{(1,1)}=-\frac{2}{5}$

Therefore,

The slope of the tangent of the curve

$=-\frac{2}{5}$

And the slope of the normal the curve

$=-\frac{1}{m}$

$=-\frac{1}{-\frac{2}{5}}$

$=\frac{5}{2}$

Hence, this is the answer.