Question

A circle with radius unity has its centre on the positive $y-axis$. If this circle touches the parabola $y=2x^2$ tangentially at the point $P$ and $Q$ then the sum of the ordinates of $P$ and $Q$ is

• A. $\frac{15}{4}$
• B. $\frac{15}{8}$
• C. $2\sqrt{15}$
• D. $5$

Answer 1

The correct option is A $\frac{15}{4}$

Let $(0,k)$ be the center of the circle:

$x^2+(y-k{)}^2=1$

P and Q both are symmetric wrt to the y-axis. hence, let $(t,2t^2)$ and $(-t,2t^2)$ be the points P and Q respectively on the parabola $y=2x^2$.

so, $t^2+(2t^2-k{)}^2=1$

Or $4t^4-t^2(4k-1)+k^2-1=0$

using quadratic formula, we have:

$t^2=\frac{4k-1\pm \sqrt{(4k-1{)}^2-16(k^2-1)}}{8}$

$(4k-1{)}^2-16(k^2-1)\ge 0$

$⟹k\le \frac{17}{8}$

in fact, $k=\frac{17}{8}$ so that we have a unique solution.

$\therefore t^2=\frac{60}{64}⟹t=\pm \frac{\sqrt{60}}{8}$

hence, the sum of the coordinates of the points is

$4t^2=\frac{60}{16}=\frac{15}{4}$