Tangents are drawn to the ellipse $\frac{x^{2}}{36} + \frac{y^{2}}{9} = 1$ from any point on the parabola $y^{2} = 4 x .$ The corresponding chord of contact will touch a parabola, whose equation is

y^{2} + 4 x = 0

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

y^{2} - 4 x = 0

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

4 y^{2} + 9 x = 0

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

y^{2} + 9 x = 0

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

Open in App

Solution

The correct option is C $4 y^{2} + 9 x = 0$
Let the tangent be drawn from a parabola $P : y^{2} = 4 x$ At point $(t^{2}, 2 t)$
We have, equation of ellipse as $E : \frac{x^{2}}{36} + \frac{y^{2}}{9} = 1$
Chord of contact from a outside point to a ellipse is
$E = 0$
$\Rightarrow \frac{x t^{2}}{36} + \frac{2 y t}{9} = 1$

Suggest Corrections

Similar questions

\frac{x^{2}}{36} + \frac{y^{2}}{9} = 1

y^{2} = 4 x

P

x^{2} + y^{2} + 4 x - 6 y + 9 {sin}^{2} α + 13 {cos}^{2} α = 0

2 α .

P

(0, 0)

x - y + 1 = 0

y^{2} = 4 x

The equation of the parabola whose focus is the point (0, 0) and the tangent at the vertex is x-y+1=0 is

Join BYJU'S Learning Program