A common tangent to $9 x^{2} - 16 y^{2} = 144$ and $x^{2} + y^{2} = 9$ is

y = \frac{2}{\sqrt{7}} x + \frac{15}{\sqrt{7}}

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

y = 3 \sqrt{\frac{2}{7} x} + \frac{15}{\sqrt{7}}

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

y = \sqrt{\frac{3}{7} x} + 15 \sqrt{7}

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

none of these

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

Open in App

Solution

The correct option is B $y = 3 \sqrt{\frac{2}{7} x} + \frac{15}{\sqrt{7}}$
Hence, Option (B) is the correct answer.

Suggest Corrections

Similar questions

9 x^{2} - 16 y^{2} = 144

x^{2} + y^{2} = 9

9 x^{2} - 16 y^{2} = 144

x^{2} + y^{2} = 9

A common tangent to $9 x^{2} - 16 y^{2} = 144 a n d x^{2} + y^{2} = 9$ is,

9 x^{2} - 16 y^{2} = 144

x^{2} + y^{2} = 9

9 x^{2} - 16 y^{2} = 144

x^{2} + y^{2} = 9

Join BYJU'S Learning Program