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Parametric Form of Tangent: Ellipse

Through a giv...

Question

Through a given point $P (h, k)$ secants are drawn to cut the circle $x^{2} + y^{2} = a^{2}$ at the points $A_{1}, B_{2}; A_{2}, B_{2}; . . . .; A_{n}, B_{n}$ .
Prove that
$(P A_{1} . P B_{1}) = (P A_{2} . P B_{2}) . . . . . \equiv (P A_{n} . P B_{n}) = P Q^{2}$
where $P Q$ is the length of tangent from $P$ to the given circle.

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Solution

$P Q^{2} = S_{1} = h^{2} + k^{2} - a^{2} . . . . (1)$
Any secant through $P (h, k)$ is $\frac{x - h}{cos θ} = \frac{y - k}{sin θ} = r$
where $r$ is the distance of $P$ from any point on the line.
$(r cos θ + h, r sin θ + k)$ is any point on the line.
If it lies on the circle, then
$(r cos θ + h)^{2} + (r sin θ + k)^{2} = a^{2}$
or $r^{2} + 2 r (h cos θ + k sin θ) + (h^{2} + k^{2} - a^{2}) = 0$
It gives two values of $r$ say $P A$ and $P B$ .
$P A . P B = r_{1} r_{2} = h^{2} + k^{2} - a^{2} = S_{2}$
$= P Q^{2} = c o n s t a n t$ , by $(1)$
as it is independent of $θ$ . Hence all the products are equal and each equals to $P Q^{2}$ .

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