Question

The radius of circle touching parabola $y^2=x$ at $(1,1)$ and having directrix of $y^2=x$ as its normal is:

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

Answer 1

The correct option is A $5$

Equation of given parabola is,

$y^2=x$

$\therefore {(y-0)}^2=4\times \frac{1}{4}(x-0)$

Comparing this equation with standard form of equation of parabola i.e. ${(y-k)}^2=4p(x-h)$, we get,

$h=0$, $k=0$ and $p=\frac{1}{4}$

Thus, vertex of the parabola is, $(h,k)$ = $O(0,0)$

Directrix of the parabola is, $x=h-p$ = $0-\frac{1}{4}=-\frac{1}{4}$

Thus, equation of directrix is $x=-\frac{1}{4}$

Let circle touches the parabola at $P(1,1)$

Let us draw tangent through this point which is tangential to both parabola and circle as shown in figure.

Thus, slope of the tangent is given by,

$m=\frac{dy}{dx}$

From equation of parabola, we can write $y=\sqrt{x}$

$\therefore m=\frac{d\left(\sqrt{x}\right)}{dx}$

$\therefore m=\frac{1}{2\sqrt{x}}$

Slope at point $(1,1)$ is given by,

$m=\frac{1}{2\sqrt{1}}$

$\therefore m=\frac{1}{2}$

Let slope of normal at point $(1,1)$ is $m_1$

$\therefore m\times m_1=-1$

$\therefore \frac{1}{2}\times m_1=-1$

$\therefore m_1=-2$

Thus, equation of normal passing through $(1,1)$ is,

$y-y_1=m_1(x-x_1)$

$y-1=-2(x-1)$

$\therefore y-1=-2x+2$

$\therefore 2x+y-3=0$

This is the equation of normal to the circle.

Now, as given in problem, directrix of parabola is normal to the circle.

Thus, we can say that equation of directrix satisfies equation of normal.

$\therefore$ put $x=\frac{-1}{4}$ in equation of normal, we get,

$2\left(\frac{-1}{4}\right)+y-3=0$

$\frac{-1}{2}+y-3=0$

$y=\frac{1}{2}+3$

$\therefore y=\frac{7}{2}$

Thus, as shown in figure, as directix is passing through center of circle, above values of x and y must represent coordinates of center of circle.

Thus, coordinates of center of circle are, $C(\frac{-1}{4},\frac{7}{2})$

Now, by distance formula,

$CP=\sqrt{{[1-\left(\frac{-1}{4}\right)]}^2+{[1-\frac{7}{2}]}^2}$

$\therefore CP=\sqrt{{[1+\frac{1}{4}]}^2+{[1-\frac{7}{2}]}^2}$

$\therefore CP=\sqrt{{\left[\frac{5}{4}\right]}^2+{\left[\frac{-5}{2}\right]}^2}$

$\therefore CP=\sqrt{\frac{25}{16}+\frac{25}{4}}$

$\therefore CP=\sqrt{\frac{25}{16}+\frac{100}{16}}$

$\therefore CP=\sqrt{\frac{125}{16}}$

$\therefore CP=\sqrt{\frac{25\times 5}{16}}$

$\therefore CP=\frac{5\sqrt{5}}{4}$

But, as shown in figure, $CP=r$

$\therefore r=\frac{5\sqrt{5}}{4}$

Thus, answer is option (C)