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  • The points on the curve 9y ^ 2 = x ^ 3, where the normal to the curve makes equal intercepts with the axes are

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 )

Answers (1)

best_answer

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)

Posted by

Gautam harsolia

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