Two tangents to a parabola meet at an angle of $45^{o}$ ; prove that the locus of their point of intersection is the curve
$y^{2} - 4 a x = (x + a)^{2}$
If they meet at an angle of $60^{o}$ , prove that the locus is
$y^{2} - 3 x^{2} - 10 a x - 3 a^{2} = 0$ .

Open in App

Solution

For the parabola $y^{2} = 4 a x$ , angle $θ$ between two tangents at points $(a t_{1}^{2}, 2 a t_{1})$ and $(a t_{2}^{2}, 2 a t_{2})$ has the relation given by $tan θ = \frac{t_{1} - t_{2}}{1 + t_{1} t_{2}}$
$∴ tan (45^{o}) = \frac{t_{1} - t_{2}}{1 + t_{1} t_{2}} = 1$
$∴ 1 + t_{1} t_{2} = t_{1} - t_{2}$
The intersection of the two tangents is given by $(a t_{1} t_{2}, a (t_{1} + t_{2}))$
$\Rightarrow h = a t_{1} t_{2}, k = a (t_{1} + t_{2})$
$\Rightarrow a (t_{1} - t_{2}) = a \sqrt{(t_{1} + t_{2})^{2} - 4 t_{1} t_{2}} = \sqrt{k^{2} - 4 a h}$
$\Rightarrow 1 + \frac{h}{a} = \frac{\sqrt{k^{2} - 4 a h}}{a}$
i.e. $k^{2} - 4 a h = (h + a)^{2}$
If the angle is $60^{o}$ , $tan (60^{o}) = \sqrt{3} = \frac{t_{1} - t_{2}}{1 + t_{1} t_{2}}$
$\Rightarrow \sqrt{3} \times (1 + \frac{h}{a}) = \frac{\sqrt{k^{2} - 4 a h}}{a}$
Squaring both sides, $3 + \frac{3 h^{2}}{a^{2}} + \frac{6 h}{a} = \frac{k^{2} - 4 a h}{a^{2}}$
$\Rightarrow 3 a^{2} + 3 h^{2} + 6 a h = k^{2} - 4 a h$
i.e. $y^{2} - 3 x^{2} - 10 a x - 3 a^{2} = 0$

Suggest Corrections

Similar questions

y^{2} = 4 a x

45^{0}

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

y^{2} = 4 a x

α

y^{2} = 4 a x

α

y^{2} = 4 a x

α

y^{2} = 4 a x

Join BYJU'S Learning Program