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Q.
The point $P$ of curve $y^2 = 2x^3$ such that the tangent at $P$ is perpendicular to the line $4x - 3y + 2 = 0$ is given by :
Application of Derivatives
Solution:
Given equation of curve : $y^{2} = 2x^{3}\,...\left(i\right)$
and line : $4x - 3y + 2 = 0$
$y^{2}=2x^{3} \Rightarrow 2y \frac{dy}{dx}=6x^{2}$
Slope of tangent
$\frac{dy}{dx}=\frac{6x^{2}}{2y}=\frac{3x^{2}}{y}=m_{1}\left(say\right)$
and $4x - 3y + 2 = 0 \Rightarrow -3y=-4x -2$
$3y=4x+2 \Rightarrow y=\frac{4}{3}x+\frac{2}{3}$
Slope of line $=\frac{4}{3}=m_{2}\left(say\right)$
Now, $m_{1}\times m_{2}=-1$
$\therefore \frac{3x^{2}}{y}\times\frac{4}{3}=-1$
$y = - 4x^{2}$
Squaring on both sides, we get
$y^{2}=16x^{4}\,...\left(ii\right)$
$equ. \left(i\right) - equ \left(ii\right) :$
$2x^{3}=16x^{4}$
$ \Rightarrow x=\frac{1}{8}$ or $0$
If $x = 0$ then $y = 0$
If $x=\frac{1}{8}$ then
$y=-4\times\frac{1}{8}\times\frac{1}{8}=-\frac{1}{16}$
Therefore, the points are $\left(0, 0\right)$ and $\left(\frac{1}{8}, -\frac{1}{16}\right).$