JEE Main 2021 — Parabola Question with Solution

Given, parabola $$y=4x^2+1$$


Let R(a, b) be mid-point of line joining point P and Q where PQ is perpendicular to line y = x.

Let coordinates of P be P(x, y), Q(q, q) and R(a, b) then,

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

Now, slope of line y = x is m₁ = 1

Slope of line PQ be

$$\frac{b-q}{a-q}=m_2$$ (say)

$$\because$$ Line y = x and PQ are perpendicular to each other,

m₁ . m₂ = $$-$$1

$$\Rightarrow \frac{b-q}{a-q}=-1\Rightarrow b-q=q-a$$

$$\Rightarrow q=\frac{b+a}{2}$$

$$\therefore$$ $$a=\frac{x+q}{2}=\frac{x+\left(\frac{b+a}{2}\right)}{2}=\frac{2x+b+a}{4}$$

$$\Rightarrow x=\frac{4a-b-a}{2}=\frac{3a-b}{2}$$

and $$b=\frac{y+q}{2}=\frac{y+\left(\frac{b+a}{2}\right)}{2}=\frac{2y+b+a}{4}$$

$$\Rightarrow y=\frac{3b-a}{2}$$

Put (x, y) in equation of parabola as P(x, y) is variable point on parabola

$$\frac{3b-a}{2}=4{\left(\frac{3a-b}{2}\right)}^2+1$$

$$\frac{(3b-a)}{2}=(3a-b{)}^2+1$$

$$\Rightarrow (3b-a)=2(3a-b{)}^2+2$$

Replace (a, b) as (x, y) $$\Rightarrow$$ (3y $$-$$ x) = 2(3x $$-$$ y)² + 2

or $$2(3x-y{)}^2+(x-3y)+2=0$$