The equation of the tangents to the ellipse $3 x^{2} + 4 y^{2} = 12$ which are parallel to the line $2 x - y + 5 = 0$ are

6 x - 2 y \pm \sqrt{\frac{155}{3}} = 0

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

2 x - y \pm \sqrt{19} = 0

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

16 x + 22 y \pm \sqrt{\frac{155}{3}} = 0

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

2 x + 2 y \pm \sqrt{39} = 0

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

Open in App

Solution

The correct option is B $2 x - y \pm \sqrt{19} = 0$
Given, ellipse $3 x^{2} + 4 y^{2} = 12$
and line $2 x - y + 5 = 0$
slope is $m = 2$
Required tangents to the ellipse $3 x^{2} + 4 y^{2} = 12$ will be of form
$y = m x \pm \sqrt{4 m^{2} + 3}$
$\Rightarrow y = 2 x \pm \sqrt{4 (2^{2}) + 3} \Rightarrow y = 2 x \pm \sqrt{19}$

Suggest Corrections

Similar questions

3 x^{2} - 4 y^{2} = 12

3 x^{2} + 4 y^{2} = 12,

y + 2 x = 4,

Equation of the tangent to $y^{2} = 16 x$ which is perpendicular to 2x-y+5=0 is

2 x + y - 6 = 0

2 x - 4 y + 7 = 0

(1, 2)

x^{2} + y^{2} - 2 x + 4 y = 0

Join BYJU'S Learning Program

Explore more

Join BYJU'S Learning Program