Two common tangents to the circle $x^{2} + y^{2} = 2 a^{2}$ and parabola $y^{2} = 8 a x$ are

x = \pm (y + 2 a)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

y = \pm (x + 2 a)

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

x = \pm (y + a)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

y = \pm (x + a)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $y = \pm (x + 2 a)$
Equation of tangent to $y^{2} = 8 a x$ is

$t y = x + 2 a t^{2} x - t y + 2 a t^{2} = 0$

It touches the given circle , so distance of centre from the tangent is equal to radius

$\Rightarrow \frac{0 - t (0) + 2 a t^{2}}{\sqrt{1 + t^{2}}} = \sqrt{2} a \Rightarrow \frac{2 a t^{2}}{\sqrt{1 + t^{2}}} = \sqrt{2} a \Rightarrow {2 a}^{2} t^{4} = a^{2} + a^{2} t^{2} \Rightarrow {2 a}^{2} t^{4} - a^{2} - a^{2} t^{2} = 0 \Rightarrow {2 a}^{2} t^{4} - 2 a^{2} t^{2} + a^{2} t^{2} - a^{2} = 0 \Rightarrow 2 a^{2} t^{2} (t^{2} - 1) + a^{2} (t^{2} - 1) = 0 \Rightarrow (2 a^{2} t^{2} + a^{2}) (t^{2} - 1) = 0 \Rightarrow (t^{2} - 1) = 0 \Rightarrow t = \pm 1$

So the equation of tangent is

$x - (\pm 1) y + 2 a {(\pm 1)}^{2} = 0 x - (\pm 1) y + 2 a = 0 \Rightarrow y = \pm (x + a)$
So option $B$ is correct

Suggest Corrections

Similar questions

x^{2} + y^{2} = 2 a^{2}

y^{2} = 8 a x

y = \pm (x + 2 a)

x^{2} + y^{2} = 2 a^{2}

y^{2} = 8 a x

2

x^{2} + y^{2} = 2 a^{2}

y^{2} = 8 a x

y^{2} = 24 x

x^{2} + y^{2} = 18

Equation of the common tangent to the parabola $y^{2} = 24 x$ and the circle $x^{2} + y^{2} = 18$ is

Join BYJU'S Learning Program