$A$ and $B$ are points of the parabola $y=x^{2}$. The tangents at $A$ and $B$ meet at $C$. The mediam of the triangle $A B C$ from $C$ has length ' $m^{\prime}$ units. Find the area of the triangle in terms of ' $m$ '.
Solution: $y=x^{2}$ $\frac{d y}{d x}=2 x$
tangent $\mathrm{BC} \Rightarrow y-x_{1}^{2}=2 x_{1}\left(x-x_{1}\right)$
$y=2 x x_{1}-x_{1}^{2}$
similarly tangent $A C$ $y=2 x x_{2}-x_{2}^{2}$
point of intersection (1) and (2) is $C$ $2 x x_{1}-x_{1}^{2}=2 x x_{2}-x_{2}^{2}$Now $\because \mathrm{CM}=m$
$\frac{x_{1}+x_{2}}{2}-x_{1} x^{2}=m$
$\frac{\left(x_{1}-x_{2}\right)^{2}}{2}=m$
$\left(x_{1}-x_{2}\right)^{2}-2 m$
$=\frac{1}{4}(2 m)^{3 / 2}=\frac{m^{3 / 2}}{\sqrt{3}}$ Ans