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Find the poin...

Question

Find the point P on the parabola $y^{2} = 4 a x$ such that area bounded by the parabola, the x-axis and the tangent at P is equal to that of bounded by the parabola , the x-axis and the normal at P

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Solution

$\begin{matrix} E q u a t i o n o f tan g e n t 2 a t y = 2 a (x + a t^{2}) t y = x + a t^{2} A_{1} = \int_{0}^{a t^{2}} \sqrt{4 a x} d x + \frac{1}{2} \times 2 a t (2 a + a t^{2} - a t^{2}) x^{3} A_{1} = 2 \sqrt{a} {[\frac{2 x^{\frac{3}{2}}}{3}]}_{0}^{a t^{2}} + 2 a^{2} t A_{1} = 2 a^{2} (\frac{2}{3} t^{2} + t) a r e a o f P Q R = 2 a^{2} t (1 + t^{2}) A_{2} = \frac{2 a^{2} t}{3} A_{1} = A_{2} 2 t^{3} + 3 t = 0 P (9 a, - 6 a) \end{matrix}$

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