Question

The point P on the parabola $y^2=4x$ for which [PR - PQ] is maximum, where $R\equiv (-2,0),Q(0,\frac{4}{3}),$ is

• A. $(1,3)$
• B. $(1,-2)$
• C. $(4,4)$
• D. $(4,-4)$

Answer 1

The correct option is C $(4,4)$

The point P on parabola $y^2=4ax$ for which $[PR-PQ]$ is maximum

Where, $R\equiv (-2,0),Q\equiv (0,\frac{4}{3})$ is

Let $(0,\frac{4}{3})$ lie outside the parabola

If $[PR-PQ]$ is maximum then PQR is the line segment

Slope of $PQ=$ Slope of $QR$

$\frac{y-\frac{4}{3}}{x-0}=\frac{\frac{4}{3}-0}{0-(-2)}\frac{3y-4}{3x}=\frac{2}{3}3y-4=2x$

Hence $(4,4)$ satisfies above equation.

Therefore point $P\equiv (4,4)$