Find the points on the curve `4x^2+9y^2=1` , where the tangents are perpendicular to the line `2y+x=0` .

Answer

Therefore points are `(frac{3}{2 sqrt{20}},-frac{1}{3 sqrt{10}})` and `(-frac{3}{2 sqrt{20}}, frac{1}{3 sqrt{10}})`

Explanation

The equation of curve is `4 x^{2}+9 y^{2}=1`.
Differentiating w.r.t `x`
$$ \begin{aligned} &8 x+18 y \frac{d y}{d x}=0 \\ &\frac{d y}{d x}=-\frac{4}{9} \frac{x}{y} \end{aligned} $$
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