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Q.
If $b, k$ are intercepts of a focal chord of the parabola $y^{2}=4 a x$ then $k$ is equal to
Conic Sections
Solution:
If coordinates of one end of focal chord are $P\left(a t^{2}, 2 a t\right)$ then the coordinates of other end will be $Q\left(\frac{a}{t^{2}}, \frac{-2 a}{t}\right)$.
$\therefore S P =\sqrt{\left(a t^{2}-a\right)^{2}+(2 a t-0)^{2}}$
$=a\left(t^{2}+1\right)=b$ (given) .... (1)
and, $S Q=\sqrt{\left(\frac{a}{t^{2}}-a\right)^{2}+\left(\frac{-2 a}{t}-0\right)^{2}}$
$=\sqrt{a\left(\frac{1}{t^{2}}+1\right)}=k $ (given) ....(2)
Dividing (1) by (2), we get $t^{2}=\frac{b}{k}$.
Putting in (1), $\frac{b}{k}+1=\frac{b}{a}$
$ \Rightarrow k=\frac{a b}{b-a}$.
