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

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Solution

Let $A B$ be the fixed line, and $O$ be the center of the fixed circle.
Taking $O$ as origin, a line through $O$ and parallel to $A B$ as axis of $y$ and a line through $O$ perpendicular to $A B$ as axis, the equation of the circle may be written as
$x^{2} + y^{2} = r^{2}$ .....(1)
And the equation of the line as
$x = a$ ....(2)
Let $P$ be any point $(h, k)$ on the locus, $P T$ be the tangent from $P$ to $1$ and $P N$ be perpendicular from $P$ upon $A B$ .
Then, $P T = \sqrt{h^{2} + k^{2} - r^{2}}$ and $P N = h - a$
$P T = P N$ or $P T^{2} = P N^{2}$
Simplifying and generalising, we get
$y^{2} = - 2 a x + a^{2} + r^{2}$ ....(3)
which is a parabola
$y^{2} = - 2 a (x - \frac{a^{2} + r^{2}}{2 a})$
Clearly the focus is ${\frac{a^{2} + r^{2}}{2 a} - \frac{a}{2}}$ or ${\frac{r^{2}}{2 a}, 0}$
And the directrix will be
$x = \frac{a^{2} + r^{2}}{2 a} + \frac{a}{2}$
$\Rightarrow 2 a x = 2 a^{2} + r^{2}$

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