Mathematics - Application of Derivative Question with Solution | TestHub

Equation of normal is $y=mx-2am-a{m}^3$ to the curve $y^2=4ax$.

$$\mathrm{a}=\frac{1}{4}$$

$\Rightarrow \mathrm{y}=\mathrm{mx}-\frac{\mathrm{m}}{2}-\frac{{\mathrm{m}}^3}{4}$ passes through $(\mathrm{c},0)$ then $-\mathrm{m}\left(\frac{{\mathrm{m}}^2}{4}+\frac{1}{2}-\mathrm{c}\right)=0$

⇒ Remaining normals are perpendicular.

⇒ Product of the roots of equation $\frac{{\mathrm{m}}^2}{4}+\frac{1}{2}-\mathrm{c}=0$ will be -1 . $\Rightarrow \frac{\frac{1}{2}-\mathrm{c}}{\frac{1}{4}}=-1\Rightarrow \mathrm{c}=\frac{1}{2}+\frac{1}{4}=\frac{3}{4}$

so $4c=3$