# Thread: Parabola Question

1. ## Parabola Question

Find the equation of the locus of the center of the circle which touches the X axis, and also touches the Circle with Center(0,3) and Radius 2 units. Also prove that this locus is a Parabola.
Please help on how to start the problem.
My approach was finding 2 points which are definitily present on the locus (by rough plot) and then proving y^2 is proportional to x, but I guess that won't work.

2. Originally Posted by Programmer
Find the equation of the locus of the center of the circle which touches the X axis, and also touches the Circle with Center(0,3) and Radius 2 units. Also prove that this locus is a Parabola.
Please help on how to start the problem.
My approach was finding 2 points which are definitily present on the locus (by rough plot) and then proving y^2 is proportional to x, but I guess that won't work.
Let M(m, t) denote the center of the circle which touches the circle around C(0,3;r=2).

1. Then the distance

$\displaystyle |\overline{CM}|=\sqrt{m^2+(t-3)^2}$

2. According to the question

$\displaystyle |\overline{CM}| - 2 = t$

3. Solve this equation for t:

$\displaystyle t=\dfrac1{10}(m^2+5)$

4. The curve which is produced by all points M is consequently:

$\displaystyle y = \dfrac1{10}(x^2+5)$

3. Originally Posted by Programmer
Find the equation of the locus of the center of the circle which touches the X axis, and also touches the Circle with Center(0,3) and Radius 2 units. Also prove that this locus is a Parabola.
Please help on how to start the problem.
My approach was finding 2 points which are definitily present on the locus (by rough plot) and then proving y^2 is proportional to x, but I guess that won't work.
Second approach:

Obviously the parabola in question is:
- opening upward
- symmetric to the y-axis

Therefore the genral equation of such a parabola is:

$\displaystyle y = ax^2+b$

There are 2 points which lay on the parabola: $\displaystyle P\left(0\ ,\ \frac12\right)$ and Q(5,3)

Plug in the coordinates of these points and solve the system of equations for a and b:

$\displaystyle \left|\begin{array}{rcl}\dfrac12&=&b\\3&=&25a+b\en d{array}\right.$

You'll get: $\displaystyle a = \dfrac1{10}~\wedge~b=\dfrac12$

4. Thanks a lot!