Question

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

• A. \((3x-y)^2+(x-3y)+2=0\)
• B. \(2(3x-y)^2+(x-3y)+2=0\)
• C. \((3x-y)^2+2(x-3y)+2=0\)
• D. \(2(x-3y)^2+(3x-y)+2=0\)

Hint:

Take \(P(p,q)\), find its foot on \(y=x\), express \(p,q\) in terms of the mid-point \((h,k)\), and substitute in \(q=4p^2+1\).

Answer 1

The Correct Option is B

Let \(P(p,q)\) with \(q=4p^2+1\). The foot of the perpendicular on \(y=x\) is \(F\left(\frac{p+q}{2},\frac{p+q}{2}\right)\).
Mid-point of PF: \(h=\frac{3p+q}{4}\), \(k=\frac{p+3q}{4}\).
Then \(3h-k=2p\Rightarrow p=\frac{3h-k}{2}\) and \(3k-h=2q\Rightarrow q=\frac{3k-h}{2}\).
Substitute in \(q=4p^2+1\): \(\frac{3k-h}{2}=(3h-k)^2+1\Rightarrow 3k-h=2(3h-k)^2+2\).
So \(2(3h-k)^2+(h-3k)+2=0\); the locus is \(2(3x-y)^2+(x-3y)+2=0\).
Answer: (2) \(2(3x-y)^2+(x-3y)+2=0\)