JEE Main 2021 — Parabola Question with Solution

Given, parabola

$$y^2=8x$$ ...... (i)

Equation of tangent at $$P(2,-4)$$ is

$$-4y=4(x+2)$$

or, $$x+y+2=0$$ ..... (ii)

and Equation of normal to the parabola is

$$x-y+C=0$$

$$\therefore$$ Normal passes through $$(2,-4)$$

$$\therefore$$ $$C=-6$$

Normal : $$x-y=6$$ ..... (iii)

Equation of directrix of parabola

$$x=-2$$ ..... (iv)

Point of intersection of tangent and normal with directrix are $$x=-2$$ at $$A(-2,0)$$ and $$B(-2,-8)$$ respectively.

$$Q(a,b)$$ and $$P(2,-4)$$ are given and AQBP is a square.

Mid-point of AB = Mid-point of PQ

$$\Rightarrow (-2,-4)=\left(\frac{a+2}{2},\frac{b-4}{2}\right)\Rightarrow a=-6,b=-4$$

$$\Rightarrow 2a+b=-16$$