Question

If equation of normal at a point $(m^2,-m^3)$ on the curve $x^3-y^2=0isy=3mx-4m^3$ then $m^2$ equals

• A. $0$
• B. $1$
• C. $-\frac{2}{9}$
• D. $\frac{2}{9}$

Answer 1

The correct option is D $\frac{2}{9}$

$x^3-y^2=0$
$\Rightarrow 3x^2-2y\left(\frac{dy}{dx}\right)=0$
$\Rightarrow 3x^2=2y\left(\frac{dy}{dx}\right)$
$\Rightarrow \frac{3x^2}{2y}=\frac{dy}{dx}$
$\Rightarrow {\left(\frac{dy}{dx}\right)}_{(m^2,-m^3)}=\frac{3\times m^4}{-2\times m^3}$
$\therefore -\left(\frac{dx}{dy}\right)=\frac{2}{3m}$
Therefore, equation of normal at given point is,
$\Rightarrow (y+m^3)=\frac{2}{3m}(x-m^2)$
$\Rightarrow 3my+3m^4=2x-2m^2$
$\Rightarrow 3my=2x-3m^4-2m^2$
$\Rightarrow y=\frac{2x}{3m}-m^3-\frac{2}{3}m..\left(1\right)$
but given equation of normal is, $y=3mx-4m^3..\left(2\right)$
Thus comparing (1) and (2), we get
$3m=\frac{2}{3m}$
$\Rightarrow m^2=\frac{2}{9}$
Hence, option 'D' is correct.