Question

Find the coordinates of the point on the curve y² =2x³ the tangent at which is perpendicular to the line 4x − 3y + 2 = 0.​

Answer 1

$\sf 4x - 3y + 2 = 0 \\ \\ \bf or \\ \\ \sf 3y = 4x + 2 \\ \\ \bf or \\ \\ \sf y = \dfrac{4}{3} x + \dfrac{2}{3}$

$\bf Hence \: slope \: of \: the \:tangent \: will \: be \\ \sf : \implies \purple{\dfrac{ - 3}{4}}$

Now,

$\sf : \implies \dfrac{dy}{dx} = \dfrac{ - 3}{4}$

$\sf : \implies y = \frac{ + }{ } \sqrt{2} . {x}^{ \dfrac{3}{2}}$

Hence,

$\sf : \implies \dfrac{dy}{dx}$

$\sf : \implies \frac{ + }{} \sqrt{2} . \dfrac{3}{2} . {x}^{ \dfrac{1}{2} }$

$\sf : \implies \dfrac{ - 3}{4} \\ \\ \sf or \\ \sf : \implies - \sqrt{2}. \dfrac{3}{2}. {x}^{ \dfrac{1}{2} } \\ \\ \\ \sf : \implies \frac{ - 3}{4} \\ \\ \sf or \\ \sf : \implies 2 \sqrt{2} . {x}^{ \dfrac{1}{2} } \\ \\ \sf : \implies 1 \\ \\ \sf or \\ \sf : \implies x = \dfrac{1}{8}$

Hence,

$\sf : \implies y = - \sqrt{2} . {x}^{ \dfrac{3}{2} } \\ \\ \\ \sf : \implies - \sqrt{2} . \frac{1}{16 \sqrt{2} } \\ \\ \\ \sf : \implies - \frac{1}{16}$

Therefore,

The co-ordinates are

$\bf : \implies \blue{( \dfrac{1}{8} \dfrac{ - 1}{16} )}$