The Touch-Me-Not Parabola!

Differentiating the equation of the parabola gives us that

$2y*\frac{dy}{dx} = 4 \Longrightarrow \frac{dy}{dx} = \frac{2}{y}$ .

Now the point on the parabola closest to the circle will clearly lie in quadrant IV where $y = -2\sqrt{x}$ , and thus where $\frac{dy}{dx} = -\dfrac{1}{\sqrt{x}}$ .

Now the least distance between the parabola and the circle will lie along the line $L$ joining the center $O(0,12)$ of the circle and the point $P(x,-2\sqrt{x})$ on the parabola at which point the tangent line is perpendicular to $L$ . Since the tangent line has slope $-\frac{1}{\sqrt{x}}$ , the slope of $L$ must then be $\sqrt{x}$ . Thus we require that

$\dfrac{-2\sqrt{x} - (-12)}{x - 0} = \sqrt{x} \Longrightarrow x\sqrt{x} + 2\sqrt{x} - 12 = 0 \Longrightarrow (\sqrt{x} - 2)(x + 2\sqrt{x} + 6) = 0$ ,

the only real solution to which is $\sqrt{x} = 2 \Longrightarrow x = 4$ .

The least distance $D$ will then be the distance between $P(4,-4)$ and $O(0,-12)$ minus the radius of the circle, and thus

$D = \sqrt{(4 - 0)^{2} + (-4 - (-12))^{2}} - 1 = \sqrt{80} - 1 = 4\sqrt{5} - 1$ .

Thus $a = 4, b = 5, c = 1$ and $a + b + c = \boxed{10}$ .

Since a = 1, the equation of parabola is y^2=4x. The parametric form of its equation would be (t^2,2t).

Now, x^2 + (y+12)^2 = 1 is a circle, whose center is (0,-12) and radius is 1

The distance between (t^2,2t) and (0,-12) is

sqrt((t^4+(2t+12)^2)

or √(t^4 +4t^2 + 48t + 144)

This distance will be minimum when (t^4 +4t^2 + 48t + 144) is minimum

i.e. its first derivative 4t^3 + 8t + 48 = 0 , the solution for is t = -2

Substituting this back in the equation for the distance: we have the shortest distance from the center of the circle to the parabola is

sqrt((-2)^4 + (12-4)^2) = sqrt( 16 + 64) = sqrt(80) = 4sqrt(5)

To get the least distance between parabola and the circle we just have to subtract the radius 1 of the circle from this.

So the distance is 4*sqrt(5) - 1

=> a = 4, b = 5 and c = 1, which tells us that a + b + c = 10

General Equation Of Normal To Parabola y = mx-2am-am^3. Pass through centre of circle

We will obtain only one real root m =2. Conormal Point(am^2 , -2am) so we get coordinate of conormal point(4,-4).

Distance Between Circle And Parabola = distance between (4,-4) and centre of circle - radius of circle