Let P be a variable point on the parabola y = 4x^2 + 1.

Question

Let P be a variable point on the parabola $y = 4{x^2} + 1$. Then, the locus of the mid-point of the point P and the foot of the perpendicular drawn from the point P to the line y = x is :

Accepted Answer

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

Solutions · 1