QJEEAdv13Ma3caa: JEE Advanced 2013 Mathematics - Class 11 JEEAdv13Ma3caa Solution: A line meets -axis at R and the arc of the parabola at the…

Given, a parabola $y^2=16 x$

Let $\mathrm{F}\left(x_0, y_0\right)=\left(4 t^2, 8 t\right)$

$\Rightarrow \quad x_0=4 t^2, y_0=8 t$

Equation of tangent at F is $t y=x+4 t^2$

G is the point of intersection of tangent $t y=x +4 t^2$ and $Y$ -axis

$\therefore \mathrm{G}=(0,4 t)$

Given, $\mathrm{G}=\left(0, y_1\right)$

$\therefore \quad y_1=4 t$

Let $\Delta$ be the area of $\Delta$ EGF

$\begin{aligned}& \Rightarrow \quad \Delta=\frac{1}{2} 4 t^2 \cdot\left(3-y_1\right) \\& \Rightarrow \quad \Delta=2 t^2(3-4 t) \\& \Rightarrow \quad \frac{d \Delta}{d t}=12 t-24 t^2=0 \text { at } t=0, \frac{1}{2} \\& \Rightarrow \frac{d^2 \Delta}{d t^2}=12-48 t<0 \text { at } t=\frac{1}{2}\end{aligned}$

Hence, $\Delta$ is maximum at $t=\frac{1}{2}$

$\therefore \quad \Delta_{\max }=\frac{1}{2}, y_0=8 t=4, y_1=4 t=2$

Given, $y=m x+3$ intersect the parabola $y^2=16 x$ at $\mathrm{F}\left(4 t^2, 8 t\right)$

$\therefore 8 t=4 m t^2+3$

Put $t=\frac{1}{2}$

$\begin{aligned}& \Rightarrow \quad 4=m+3 \\& \Rightarrow m=1\end{aligned}$

Hints :

(i) $\frac{d y}{d x}=0$ at $x=\alpha$ and \frac{d^2 y}{d x^2}<0 at $x=\alpha$ , then the function has point of local maxima at $x=a$ .

(ii) The parametric point of the parabola $y^2= 4 a x$ is $\left(a t^2, 2 a t\right)$

(iii) The equation of tangent of parabola $y^2=4 a x$ at $\left(a t^2, 2 a t\right)$ is $t y=x+a t^2$ .