Find the equation of the hyperbola in the standard form if: Length of

1.	Find the equation of the hyperbola in the standard form if: Length of conjugate axis is 5 and distance between foci is 13.
Answer» Let the required equation of hyperbola be \(\frac {x^2}{a^2} - \frac {y^2}{b^2} = 1\) Length of conjugate axis = 2b Given, length of conjugate axis = 5 2b = 5 b = 5/2 b² = 25/4 Distance between foci = 2ae Given, distance between foci = 13 2ae = 13 ae = 13/2 a² e² = 169/4 Now, b² = a² (e² – 1) b² = a² e² – a² 25/4 = 169/4 - a² a² = 169/4 - 25/4 = 36 ∴ The required equation of hyperbola is \(\frac {x^2}{36} - \frac {y^2}{\frac {25}{4}} = 1\) i.e., \(\frac {x^2}{36} - \frac {4y^2}{25} = 1\)

Find the equation of the hyperbola in the standard form if: Length of conjugate axis is 5 and distance between foci is 13.

Answer»

Let the required equation of hyperbola be \(\frac {x^2}{a^2} - \frac {y^2}{b^2} = 1\)

Length of conjugate axis = 2b

Given, length of conjugate axis = 5

2b = 5

b = 5/2

b² = 25/4

Distance between foci = 2ae

Given, distance between foci = 13

2ae = 13

ae = 13/2

a² e² = 169/4

Now, b² = a² (e² – 1)

b² = a² e² – a²

25/4 = 169/4 - a²

a² = 169/4 - 25/4 = 36

∴ The required equation of hyperbola is \(\frac {x^2}{36} - \frac {y^2}{\frac {25}{4}} = 1\)

i.e., \(\frac {x^2}{36} - \frac {4y^2}{25} = 1\)