The eccentricity of the hyperbola $5 x^{2} - 4 y^{2} + 20 x + 8 y = 4$ is

\sqrt{2}

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

\frac{3}{2}

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

2

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

3

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

Open in App

Solution

The correct option is C $\frac{3}{2}$
Given hyperbola is $5 x^{2} - 4 y^{2} + 20 x + 8 y = 4$
$\Rightarrow 5 (x^{2} + 4 x) - 4 (y^{2} - 2 y) = 4$
$\Rightarrow 5 (x^{2} + 4 x + 4) - 20 - 4 (y^{2} - 2 y + 1) + 4 = 4$
$\Rightarrow 5 (x + 2)^{2} - 4 (y - 1)^{2} = 20$
$\Rightarrow \frac{(x + 2)^{2}}{4} - \frac{(y - 1)^{2}}{5} = 1$
Now, $e = \sqrt{1 + \frac{b^{2}}{a^{2}}}$
$∴ e = \sqrt{\frac{4 + 5}{4}} = \sqrt{\frac{9}{4}} = \frac{3}{2}$

Suggest Corrections

Similar questions

5 x^{2} - 4 y^{2} - 20 x + 8 y + 4 = 0

4 x^{2} - y^{2} - 8 x - 8 y - 28 = 0

5 x^{2} - 4 y^{2} - 30 x - 8 y - 39 = 0

Let the eccentricity of the hyperbola

be reciprocal to that of the ellipse x²+4y²=4. If the hyperbola passes through a focus of the ellipse, then

9 x^{2} - 16 y^{2} + 72 x - 32 y - 16 = 0

4 x^{2} - 5 y^{2} - 16 x + 10 y + 31 = 0

5 x^{2} - 4 y^{2} + 20 x + 8 y - 4 = 0

4 (y + 3)^{2} - 9 (x - 2)^{2} = 1

4 x^{2} - 9 y^{2} - 8 x - 32 = 0

Join BYJU'S Learning Program