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

Answer

Slope of the tangent `=frac{-2}{5}`


Slope of the normal `= frac{5}{2}`

Explanation

ATQ,

`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 \\
&\Rightarrow \frac{d y}{d x}(3+2 y)=-2 x \\
&\Rightarrow \frac{d y}{d x}=\frac{-2 x}{3+2 y}
...