Find the locus of the middle points of chords of the parabola which subtend a constant angle a at the vertex.

Open in App

Solution

For the parabola $y^{2} = 4 a x$ , chord joining points $(a t_{1}^{2}, 2 a t_{1})$ and $(a t_{2}^{2}, 2 a t_{2})$ has the equation $y - 2 a t_{2} = \frac{2 a t_{2} - 2 a t_{1}}{a t_{2}^{2} - a t_{1}^{2}} \times (x - a t_{2}^{2})$
i.e. $(y - 2 a t_{2}) (t_{1} + t_{2}) = 2 (x - a t_{2}^{2})$
Since the chord subtends angle $a$ at the vertex, $\frac{\frac{2}{t_{2}} - \frac{2}{t_{1}}}{1 + \frac{2}{t_{2}} \times \frac{2}{t_{1}}} = tan a$
$∴ 2 (t_{1} - t_{2}) = tan a \times (t_{1} t_{2} + 4) . . . (1)$
The midpoint of the chord is given by $(\frac{a (t_{1}^{2} + t_{2}^{2})}{2}, a (t_{1} + t_{2}))$
Let the midpoint be denoted by $(x, y)$
$∴ 2 x = a (t_{1}^{2} + t_{2}^{2}), y = a (t_{1} + t_{2})$
Equation (1) becomes $2 \times \sqrt{\frac{y^{2}}{a^{2}} - 4 (\frac{y^{2}}{2 a^{2}} - \frac{2 x}{2 a})} = tan a \times (\frac{y^{2}}{2 a^{2}} - \frac{2 x}{2 a} + 4)$
$\Rightarrow \frac{2}{a} \times \sqrt{y^{2} - 2 y^{2} + 4 a x} = tan a \times \frac{y^{2} - 2 a x + 8 a^{2}}{2 a^{2}}$
$\Rightarrow 4 a \sqrt{4 a x - y^{2}} = tan a \times (y^{2} - 2 a x + 8 a^{2})$
Or $16 a^{2} (4 a x - y^{2}) = {tan}^{2} a \times (y^{2} - 2 a x + 8 a^{2})^{2}$ is the required locus

Suggest Corrections

Similar questions

y^{2} = 4 a x

y^{2} = 4 a x

α

y^{2} = 4 a x

y^{2} = 2 a (x - 4 a)

P Q

t_{1}

t_{2}

t_{1} t_{2} = - 2

Join BYJU'S Learning Program