Question

Consider the parabola with vertex $(\frac{1}{2},\frac{3}{4})$ and the directrix $y=\frac{1}{2}$. Let $P$ be the point where the parabola meets the line $x=-\frac{1}{2}$. If the normal to the parabola at $P$ intersects the parabola again at the point $Q$, then $(PQ{)}^2$ is equal to

• A. $\frac{15}{2}$
• B. $\frac{125}{16}$
• C. $\frac{75}{8}$
• D. $\frac{25}{2}$

Answer 1

The correct option is B $\frac{125}{16}$


The equation of parabola is
${(x-\frac{1}{2})}^2=(y-\frac{3}{4})$
$\therefore y=x^2-x+1$
$\therefore \text{Point P}=(-\frac{1}{2},\frac{7}{4})$
$\therefore \frac{dy}{dx}=2x-1$
Slope of normal at $x=-\frac{1}{2}\text{is}\frac{1}{2}.$
Equation of normal is : $y-\frac{7}{4}=\frac{1}{2}(x+\frac{1}{2})$
$\therefore 2x-4y+8=0$
$\therefore x-2y+4=0$
$\therefore \text{Coordinate of Q}=(2,3)$
$\therefore P{Q}^2={(2+\frac{1}{2})}^2+{(3-\frac{7}{4})}^2=\frac{125}{16}$