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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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Safeer PP

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