Two parabolas have the same axis and tangents are drawn to the second from points on the first; prove that the locus of the middle points of the chords of contact with the second parabola all lie on a fixed parabola.

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Solution

$y^{2} = 4 a x . . . . (1)$
$y = 4 b (x + c) . . . . (2)$
be the equation of the parabolas and let PQ be the polar, the co ordinates of ends P and Q to be $(a t_{1}^{2}, 2 a t_{1})$ and $(a t_{1}^{2}, 2 a t_{2})$ respectively; then the equation of polar PQ is
$(t_{1} + t_{2}) y = 2 x + 2 a t_{1} t_{2} . . . . (3)$
and if $(x_{1}, y_{1})$ be the pole, then polar of $(x_{1}, y_{1})$ w.r.t. $(1)$ is
$y y_{1} = 2 a (x + x_{1}) . . . . (4)$
Since $(3)$ and $(4)$ represents the same line, the co efficient will be proportional. So, comparing the co efficients of x,y and constant term, we get
$\frac{y_{1}}{t_{1} + t_{2}} = \frac{2 a}{2} = \frac{2 a x_{1}}{2 a t_{1} t_{2}}$
or $x = a t_{1} t_{2}$ and $y = a (t_{1} + t_{2})$
Hence the co ordinates of the pole are
${a t_{1} t_{2}, a (t_{1} + t_{2})}$
At this point lies on $(2)$ , we have
$a^{2} {(t_{1} + t_{2})}^{2} = 4 b (a t_{1} t_{2} + c)$
$= \frac{2 b (2 a^{2} t_{1} t_{2} + 2 a c)}{a} . . . . . (5)$
If $(h, k)$ be the co ordinates of the middle point of the chord PQ then
$\frac{2 h}{a} = t_{1}^{2} + t_{2}^{2}$ and $\frac{k}{a} = t_{1} + t_{2}$
Hence $k^{2} - 2 a h = 2 a t_{1} t_{2}$
Substituting the values in $(5)$
$a^{2} (\frac{k^{2}}{a^{2}}) = (\frac{2 b}{a}) (k^{2} - 2 a h + 2 a c)$
Generalizing , we have $a y^{2} = 2 b (y^{2} - 2 a x + 2 a c)$ which is a fixed parabola

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