Question

If the tangent to the curve $y^2=x^3$ at $(m^2,m^3)$ is also a normal to the curve at $(M^2,M^3),$ then the value of $mM$ is

• A. $-\frac{1}{9}$
• B. $-\frac{2}{9}$
• C. $-\frac{1}{3}$
• D. $-\frac{4}{9}$

Answer 1

The correct option is D $-\frac{4}{9}$

$y^2=x^3\Rightarrow 2y\frac{dy}{dx}=3x^2$
Slope, $m=\frac{3x^2}{2y}$

Slope of tangent at $(m^2,m^3)$
$\Rightarrow m_1=\frac{3m^4}{2m^3}$
$\Rightarrow m_1=\frac{3m}{2}$

Slope at $(M^2,M^3)$
$m_2=\frac{3M}{2}$

Now, $m_1{m}_2=-1$
$\Rightarrow \frac{3m}{2}\cdot \frac{3M}{2}=-1\Rightarrow mM=-\frac{4}{9}$