Find the points on the curve `9y^2=x^3` where normal to the curve makes equal intercepts with the axes.

Answer

points are `(4, 8 / 3)` and `(4, -8 / 3)`

Explanation

We know that Normal makes equal intercept
$$ \left(M_{N}\right) \text { slope }=\pm 1=\frac{-1}{dy / dx} $$
ATQ curve is `9 y^{2}=x^{3}`

Now when differentiating,


`18 y \frac{d y}{d x}=3 x^{2} \Rightarrow \frac{d y}{d x}=\frac{x^{2}}{6 y}=\pm 1`

Squaring `x^{4}=36 y^{2}`


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