A quadrilateral is inscribed in a parabola $y^{2} = 4 a x$ and three of its sides pass through fixed points on the axis. Show that the fourth side also passes through a fixed point on the axis of the parabola.

Open in App

Solution

Slop of $A B = \frac{2 a t_{2} - 2 a t_{1}}{a t_{2}^{2} - a t_{1}^{2}} = \frac{3 a (t_{2} - t_{1})}{a (t_{2}^{2} - t_{1}^{2})}$
$= \frac{2}{t_{1} + t_{2}}$
equation chord AB
$y - y_{1} = m (x - x_{1})$
$y - 2 a t_{1} = \frac{2}{t_{1} + t_{2}} (x - a t_{1}^{2})$
$(t_{2} + t_{1}) y - 2 a t_{1} (t_{2} + t_{1}) = 2 x - 2 a t_{1}^{2}$
$(t_{1} + t_{2}) y - 2 a t_{1} t_{2} - 2 a t_{1}^{2} = 2 x - 2 a t_{1}^{2}$
$2 x - (t_{1} + t_{2}) y + 2 a t_{1} t_{2} = 0$ ___(1)
(1) line bases through $(c, 0)$
$2 c - (t_{1} + t_{2}) \times 0 + 2 a t_{1} t_{2} = 0$
$t_{1} t_{2} = \frac{- c}{a}$ ____(1)
for chord $B C = t_{2} t_{3} = - \frac{d}{a} (d, 0)$ __(2)
for chord $C D = t_{3} t_{4} = - \frac{e}{a} (e, 0)$ ___(3)
$(1) \times (3)$
$t_{1} t_{2} t_{3} t_{4} = - \frac{c}{a} . - \frac{e}{a} = \frac{e c}{a^{2}}$
$t_{1} t_{4} (- \frac{d}{a}) = \frac{e c}{a^{2}}$
$t_{1} t_{4} = \frac{e c}{a^{2} a} - \frac{c}{a} = - \frac{e c}{a d} = - \frac{e c}{a d}$
Hence chord AD based through fixed point.

Suggest Corrections

Similar questions

y^{2} = 4 a x

(h, k)

y^{2} = 4 a x

(h, k)

y^{2} = 4 a x

P (α^{2}, k_{1})

Q (k_{2}, - \frac{16}{α})

y^{2} = 16 x

A

y^{2} = 4 a x

A P

A Q

P Q

Join BYJU'S Learning Program