Question: If tangent to the curve y2 = x3 at point (m2, m3) is als...

Question

If tangent to the curve y² = x³ at point (m², m³) is also a normal to the curve at point (M², M³), then mM is equal to –

Answer 1

y² = x³ differentiating Ž 2y. $\frac{dy}{dx}$ = 3x²

Ž $\left( \frac{dy}{dx} \right)_{(m^{2},m^{3})}$ = $\frac{3x^{2}}{2y}$ = $\frac{3m^{4}}{2m^{3}}$ = $\frac{3}{2}$ m

Slope of normal = $\frac{–1}{\left( \frac{dy}{dx} \right)}_{(M^{2},M^{3})}$ = – $\left( \frac{2y}{3x^{2}} \right)$

= – $\frac{2M^{3}}{3M^{4}}$ = – $\frac{2}{3M}$

Ž $\frac{3}{2}$ m = – $\frac{2}{3M}$ Ž $\begin{matrix} mM = –\frac{4}{9} \end{matrix}$

Question: If tangent to the curve y<sup>2</sup> = x<sup>3</sup> at point (m<sup>2</sup>, m<sup>3</sup>) is als...