Find the slopes of the tangent and the normal to the curve `x^2+3y y2=5` at `(1,1)`

Answer

Slope at tangent `=frac{-2}{5}`


Slope at normal `=frac{5}{2}`

Explanation

ATQ, Eq of the curve is
$$ x^{2}+3 y+y^{2}=5 $$
On differentiating both sides w.r.t. `x` , we get
$$
\begin{aligned}
&2 x+3 \frac{d y}{d x}+2 y \frac{d y}{d x}=0 \\
...