QJEEM21Maffaf: JEE Main 2021 Mathematics - Class 11 JEEM21Maffaf Solution: Let P be a variable point on the parabola . Then, the locus of…

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


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,

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

Now, slope of line y = x is m₁ = 1

Slope of line PQ be

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

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

m₁ . m₂ = $-$ 1

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

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

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

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

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

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

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

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

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

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

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

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