Question

If the tangent at any point $P \left(4 m ^{2}, 8 m ^{3}\right)$ of $x ^{3}- y ^{2}=0$ is also a normal to the curve $x ^{3}- y ^{2}=0$, then the value of $m$ is

• A. $m =\frac{\sqrt{2}}{3}$
• B. $m=-\frac{\sqrt{2}}{3}$
• C. $m =\frac{3}{\sqrt{2}}$
• D. $m =-\frac{3}{\sqrt{2}}$

Answer 1

Answer: (A), (B)

$x^{3}-y^{2}=0 \,\,\, ...(i)$
$\Rightarrow 2 y \times \frac{d y}{d x}=3 x^{2}$
Slope of the tangent at $P=\left.\frac{d y}{d x}\right|_{p}=\left.\frac{3 x^{2}}{2 y}\right|_{\left(4 m^{2} .8 m^{3}\right)}=3 m$
$\therefore $ Equation of the tangent at $P$ is
$y-8 m^{3}=3 m\left(x-4 m^{2}\right)$
or $y=3 m x-4 m^{3}\,\,\, ...(ii)$
It cuts the curve again at point $Q$.
Solving (i) and (ii), we get $x =4 m ^{2}, m ^{2}$
Put $x=m^{2}$ in equation (ii)
$\Rightarrow y=3 m\left(m^{2}\right)-4 m^{3}=-m^{3}$
$\therefore Q$ is $\left(m^{2},-m^{3}\right)$
Slope of the tangent at $Q =\left.\frac{ dy }{ dx }\right|_{\left( m ^{2},- m ^{3}\right)}=\frac{3\left( m ^{4}\right)}{2 \times\left(- m ^{3}\right)}=\frac{-3}{2} m$
Slope of the normal at $Q =\frac{1}{(-3 / 2) m }=\frac{2}{3 m }$
Since tangent at $P$ is normal at $Q $
$\Rightarrow \frac{2}{3 m }=3 m $
$\Rightarrow 9 m ^{2}=2$