The equation of the normal to the hyperbola $x^{2} - 4 y^{2} = 5$ at (3, -1) is

4x + 3y – 15 = 0

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

4x – 3y – 15 = 0

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

4x – 3y + 5 = 0

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

4x + 4y + 15 = 0

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

Open in App

Solution

The correct option is A 4x + 3y – 15 = 0
Equation of the tangent at $(3, - 1) i s S_{1} = 0 \Rightarrow 3 x + 4 y = 5$
$∴$ Equation of the normal is a line perpendicular to 3x + 4y = 5 and passing (3, -1)
i.e, 4x – 3y = 15

Suggest Corrections

Similar questions

x^{2} - 4 y^{2} = 36

(10, 4)

\frac{x^{2}}{16} - \frac{y^{2}}{9} = 1

(- 4, 0)

\frac{x^{2}}{25} - \frac{y^{2}}{16} = 1

(5, 0)

x^{2} + 4 y^{2} = 5

\frac{x^{2}}{16} - \frac{y^{2}}{9} = 1

(6, \frac{3}{2} \sqrt{5})

Join BYJU'S Learning Program