Circle in a Parabola

Equation of normal at point $(t,t^2)$ is $y= - \frac{x}{2t} + \frac{1}{2} +t^2$ .

The normal passes through the centre of the circle, which lies on the $y-axis$ .

So, the centre is $(0, \frac{1}{2} +t^2)$ .

But the distance of the line joining $(t,t^2)$ and $(-t,t^2)$ from the $y-axis$ is $t^2$ .

So, the required distance is $\frac{1}{2} +t^2-t^2= \frac{1}{2}$ .

The line tangent to the parabola at point $P=(x, x^2)$ has slope $2x$ , so the line perpendicular to the parabola at this point has slope $\frac{-1}{2x}$ . This perpendicular line meets the $y$ axis at point $Q$ such that $\frac{y_Q-y_P}{x_Q-x_P} = \frac{-1}{2x}$ . But $x_Q-x_P=-x$ , which gives $y_Q-y_P=\frac{1}{2}$ .

I know this is in the calculus section, but here's a solution without calculus.

Let the equation of the circle be $x^2 + (y-a)^2 = b^2$

The center of the circle would then be $(0,a)$

Substituting $y =x^2$ , $y + (y-a)^2 = b^2$

$y^2 -(2a -1)y +(a^2 - b^2) = 0$

Using the quadratic formula,

$y = \frac{ (2a-1) \pm \sqrt {(2a -1)^2 - 4(a^2 -b^2)}}{2}$

Since it's symmetrical, there is only one solution and $(2a -1)^2 - 4(a^2 -b^2) = 0$

$y = \frac { 2a -1}{2} = a - \frac{1}{2}$

Therefore, height difference = $\boxed{0.5}$

Eqn. of circle with center (0,a) is x^2 + (y-a)^2 = r^2. At the point of tangency slopes of two curves are equal: ==> -x/(y-a) = 2x ==> (a-y) = 1/2