JEE Main 2022 — Parabola Question with Solution

$$\tan \frac{\pi }{4}=\left|\frac{m-3}{1+3m}\right|$$

$$\Rightarrow \frac{m-3}{1+3m}=\pm 1$$

$$\Rightarrow \frac{m-3}{1+3m}=+1$$ and $$\frac{m-3}{1+3m}=-1$$

$$\Rightarrow m-3=1+3m$$ and $$m-3=-1-3m$$

$$\Rightarrow 2m=-4$$ and $$4m=2$$

$$m=-2$$ and $$m=\frac{1}{2}$$

We know, Equation of tangent to the parabola $$y^2=4m$$ is $$y=mx+\frac{a}{m}$$ and point of contact is $$\left(\frac{a}{m^2},\frac{2a}{m}\right)$$

$$\therefore$$ Equation of tangent

$$y=-2x-\frac{a}{2}$$

and $$y=\frac{x}{2}+2a$$

$$\therefore$$ Point of contact A and B are

$$A\left(\frac{a}{{(-2)}^2},\frac{2a}{-2}\right)=A\left(\frac{a}{4},-a\right)$$

$$B\left(\frac{a}{{\left(\frac{1}{2}\right)}^2},\frac{2a}{\left(\frac{1}{2}\right)}\right)=B\left(4a,4a\right)$$

As points A, B and S are colinear so area of triangle formed by those 3 points are zero.

Area of $$\Delta$$ABS = {1 \over 2}\left| {\matrix{ {{a \over 4}} & { - a} & 1 \cr {4a} & {4a} & 1 \cr a & 0 & 1 \cr } } \right|

$$=\frac{a}{4}(4a-0)+a(4a-a)+1(0-4a^2)$$

$$=a^2+3a^2-4a^2=0$$

$$\therefore$$ Area of triangle is independent of value of a.

So, for all value of a > 0 (already given a must be greater than 0) point A, B and S will be collinear.