# A point moves so that its distance from a fixed line equals the length of the tangent segment to a given circle. Show that the locus is a parabola.

What must be the approach to solve the following question?

A point moves so that its distance from a fixed line is equal to the length of the tangent drawn from it to a given circle. Prove that the locus is a parabola. Find the position of focus and directrix.

On which parameter does the equation of parabola depend?

• Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments. – José Carlos Santos Sep 25 '18 at 7:21

## 1 Answer

You have a line and a circle. You can always move and rotate the coordinate system so that $$x$$ axis represents the line and the circle has its center on the $$y$$ axis. Let us assume that circle has radius $$R$$ and its center $$C$$ has coordinates $$(0,d)$$.

Moving point $$M$$ has coordinates $$(x,y)$$. Obviously, the distance from the line is defined by it's $$y$$ coordinate. Now draw a tangent from point $$M$$ to the circle and denote the touching point with $$T$$.

Obviously:

$$CM^2=MT^2+TC^2$$

or:

$$x^2+(y-d)^2=R^2+y^2$$

$$x^2+y^2-2yd+d^2=R^2+y^2$$

$$x^2+d^2-R^2=2yd$$

$$\frac1{2d}(x^2+d^2-R^2)=y$$

And this is obviously the equation of parabola with $$y$$ axis being its axis of symmetry.

It's a well known fact that for parabola $$y=a(x−h)^2+k$$, the focus is at $$(h,k+\frac{1}{4a})$$. In this example $$a=\frac1{2d}$$, $$h=0$$, $$k=\frac{d^2-R^2}{2d}$$.