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If the tangent to the curve $y^{2}=x^{3}$ at $\left(m^{2}, m^{3}\right)$ is also a normal to the curve at $\left(M^{2}, M^{3}\right)$, then the value of $m M$ is
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Verified Answer
The correct answer is:
$-\frac{4}{9}$
Hint:
$2 y y_{1}=3 x^{2}$
$y_{1}=\frac{3 x^{2}}{2 y} \Rightarrow\left(y_{1}\right)_{m^{2}, m^{3}}=\frac{3 \times m^{4}}{2 \times m^{3}}=\frac{3 m}{2}$
Again; slope of normal $=-\frac{2}{3 M}, m M=-\frac{4}{9}$
$2 y y_{1}=3 x^{2}$
$y_{1}=\frac{3 x^{2}}{2 y} \Rightarrow\left(y_{1}\right)_{m^{2}, m^{3}}=\frac{3 \times m^{4}}{2 \times m^{3}}=\frac{3 m}{2}$
Again; slope of normal $=-\frac{2}{3 M}, m M=-\frac{4}{9}$
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