# 24) The points on the curve  $9 y^2 = x ^3$, where the normal to the curve makes equal intercepts with the axes are$A ) \left ( 4 , \pm \frac{8}{3} \right )\\\\ .\: \: \: \: \: B ) \left ( 4 , \frac{-8}{3} \right ) \\\\ . \: \: \: \: \: C) \left ( 4 , \pm \frac{3}{8} \right ) \\\\ . \: \: \: \: D ) \left ( \pm 4 , \frac{3}{8} \right )$

Given the equation of the curve
$9 y^2 = x ^3$
We know that the slope of the tangent at a point on a given curve is given by  $\frac{dy}{dx}$
$18y\frac{dy}{dx} = 3x^2\\ \frac{dy}{dx} = \frac{x^2}{6y}$
We know that
$Slope \ of \ normal = \frac{-1}{Slope \ of \ tangent } = \frac{-1}{\frac{x^2}{6y}} = \frac{-6y}{x^2}$
At point (a,b)
$Slope = \frac{-6b}{a^2}$
Now, the equation of normal with point (a,b) and $Slope = \frac{-6b}{a^2}$

$y-y_1=m(x-x_1)\\ y-b=\frac{-6b}{a^2}(x-a)\\ ya^2 - ba^2 = -6bx +6ab\\ ya^2+6bx=6ab+a^2b\\ \frac{y}{\frac{6b+ab}{a}}+\frac{x}{\frac{6a+a^2}{6}} = 1$
It is given that  normal to the curve makes equal intercepts with the axes
Therefore,
$\frac{6b+ab}{a}=\frac{6a+a^2}{6} \\ 6b(6 + a) =a^2( 6+a)\\ a^2 = 6b$
point(a,b) also satisfy the given equation of the curve
$9 b^2 = a ^3\\ 9(\frac{a^2}{6})^2 = a^3\\ 9.\frac{a^4}{36} = a^3\\ a = 4$
$9b^2 = 4^3\\ 9b^2 =64\\ b = \pm\frac{8}{3}$
Hence, The points on the curve  $9 y^2 = x ^3$, where the normal to the curve makes equal intercepts with the axes are $\left ( 4,\pm\frac{8}{3} \right )$
Hence, the correct answer is (A)

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