The sides of a triangle touch a parabola, and two of its angular points lie on another parabola with its axis in the same direction; prove that the locus of the third angular point is another parabola.

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Solution

Let the point of contact of sides of triangle be $P (a t_{1}^{2}, 2 a t_{1}), Q (a t_{2}^{2}, 2 a t_{2})$ and $R (a t_{3}^{2}, 2 a t_{3})$

Two of the coordinates of point of intersection of the sides of triangle are

$B (a t_{2} t_{3}, a (t_{2} + t_{3})), C (a t_{3} t_{1}, a (t_{3} + t_{1}))$

Let they lie on the parabola ${(y - k)}^{2} = 4 b (x - h)$

Substituting $B$ and $C$ in the equation of parabola

${(y - a (t_{2} + t_{3}))}^{2} = 4 b (x - a t_{2} t_{3}) . . . . . . . . . (i) {(y - a (t_{3} + t_{1}))}^{2} = 4 b (x - a t_{3} t_{1}) . . . . . . . . . . . (i i)$

Let the third angular point be $(x, y)$

$x = a t_{1} t_{2} . . . . . . (i i i) y = a (t_{1} + t_{2}) . . . . . . . . . (i v)$

We need to eliminate

Subtracting $(i i)$ from $(i)$

${(y - a (t_{2} + t_{3}))}^{2} - {(y - a (t_{3} + t_{1}))}^{2} = 4 b (x - a t_{2} t_{3}) - 4 b (x - a t_{3} t_{1}) a ((t_{1} + t_{2}) + 2 a t_{3} - 2 k = 4 b t_{3}$

using $(i v)$

$y + 2 a t_{3} - 2 k = 4 b t_{3} t_{3} = \frac{y - 2 k}{4 b - 2 a} . . . . . . . . . . . . . . . (v)$

Mutiplying $(i)$ by $t_{1}$ and $(i i)$ by $t_{2}$ and subtracting

$\Rightarrow - a^{2} t_{1} t_{2} + a^{2} t_{3}^{2} + k^{2} - 2 a k t_{3} = - 4 b h$

substituting $(i i i)$ and $(v)$

$- a x + a^{2} {(\frac{y - 2 k}{4 b - 2 a})}^{2} + k^{2} - 2 a k (\frac{y - 2 k}{4 b - 2 a}) = - 4 b h {(y - 2 k)}^{2} - 2 a k (y - 2 k) (4 b - 2 a) = {(4 b - 2 a)}^{2} (a x - 4 b h - k^{2})$

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