The eccentricity of the hyperbola $5 x^{2} - 4 y^{2} - 20 x + 8 y + 4 = 0$ is ______________________________.

Open in App

Solution

Given hyperbola is,

$5 x^{2} - 4 y^{2} + 20 x + 8 y + 4 = 0 i . e . 5 (x^{2} + 4 x) - 4 (y^{2} - 2 y) + 4 = 0 i . e . 5 (x^{2} + 4 x + 4 - 4) - 4 (y^{2} - 2 y + 1 - 1) + 4 = 0 i . e . 5 [{(x + 2)}^{2} - 4] - 4 [{(y - 1)}^{2} - 1] + 4 = 0 i . e . 5 {(x + 2)}^{2} - 4 {(y - 1)}^{2} - 20 + 4 + 4 = 0 i . e . 5 {(x + 2)}^{2} - 4 {(y - 1)}^{2} - 12 = 0 i . e . \frac{5 {(x + 2)}^{2}}{12} - \frac{{(y - 1)}^{2}}{3} = 0 i . e . a^{2} = (\frac{12}{5}) and b^{2} = 3$

Suggest Corrections

Similar questions

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

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

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

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

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

Join BYJU'S Learning Program