Find the points on the curve 9y^{2} = x^{3} where the normal to the curve makes equal intercepts with both the axes. Also find the equation of the normals

 

Answers (1)

Curve: 9y^{2} = x^{3}

Differntiate w.r. x

\\ 18{y} \frac{{dy}}{{dx}}=3 {x}^{2} \\\\ \frac{{dy}}{{dx}}=\frac{{x}^{2}}{6 {y}} \\

Slope of the normal 

\\ -\frac{{dx}}{{dy}}=-\frac{6 {y}}{{x}^{2}} \\

Given slope of normal is \\ \pm 1 \\.

-\frac{6 {y}}{{x}^{2}} = \pm 1

y = \pm \frac{x^{2}}{6}

For the point of contacts 

9y^{2} = x^{3}

9 \frac{x^{4}}{36}=x^{3}

x^{3}(x-4)=0

x=0,4

(x=0 \text { is not possible })

x = 4

y^{2}=\frac{64}{9}

y=\pm \frac{8}{3}

\\\text { Point of contacts: } \\ \\\left(4, \frac{8}{3}\right),\left(4, \frac{-8}{3}\right)

$ Equation of normal at $\left(4, \frac{8}{3}\right)$ : $

\\ y-\frac{8}{3}=-(x-4) \\ \\ 3 x+3 y=20

$ Equation of normal at $\left(4, -\frac{8}{3}\right)$ : $

\\ y+\frac{8}{3}=(x-4) \\ \\ 3 x-3 y=20

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