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

Answer

Slope of tangent at `(1,1)` is `-2 / 5`

Slope of normal`= 5 / 2`

Explanation

Given as `x^{2}+3 y+y^{2}=5` at `(1,1)`

Differentiate the above equation with respect to `x`
$$ \Rightarrow \frac{d}{d x}\left(x^{2}+3 y+y^{2}\right)=\frac{d}{d x}(5) $$
$$ \Rightarrow \frac{d}{d x}\left(x^{2}\right)+\frac{d}{d x}(3 y)+\frac{d}{d x}\left(y^{2}\right)=\frac{d}{d x (5) } $$
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