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Somnath Sumit Kabra
Asked: 2022-11-09T22:37:00+05:30 In:

Find the equation of the hyperbola whose foci are (± 5, 0) and the conjugate axis is of length 8 ?
1. \(\frac{{{x^2}}}{{{9}}} – \frac{{{y^2}}}{{{16}}} = 1\)
2. \(\frac{{{x^2}}}{{{9}}} – \frac{{{y^2}}}{{{7}}} = 1\)
3. \(\frac{{{x^2}}}{{{25}}} – \frac{{{y^2}}}{{{16}}} = 1\)
4. None of these

Find the equation of the hyperbola whose foci are (± 5, 0) and the conjugate axis is of length 8 ?
1. \(\frac{{{x^2}}}{{{9}}} – \frac{{{y^2}}}{{{16}}} = 1\)
2. \(\frac{{{x^2}}}{{{9}}} – \frac{{{y^2}}}{{{7}}} = 1\)
3. \(\frac{{{x^2}}}{{{25}}} – \frac{{{y^2}}}{{{16}}} = 1\)
4. None of these
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  1. Correct Answer – Option 1 : \(\frac{{{x^2}}}{{{9}}} – \frac{{{y^2}}}{{{16}}} = 1\)

    CONCEPT:

    The properties of a rectangular hyperbola \(\frac{{{x^2}}}{{{a^2}}} – \frac{{{y^2}}}{{{b^2}}} = 1\) are:

    • Its centre is given by: (0, 0)
    • Its foci are given by: (- ae, 0) and (ae, 0)
    • Its vertices are given by: (- a, 0)  and (a, 0)
    • Its eccentricity is given by: \(e = \frac{{\sqrt {{a^2} + {b^2}} }}{a}\)
    • Length of transverse axis = 2a and its equation is y = 0.
    • Length of conjugate axis = 2b and its equation is x = 0.
    • Length of its latus rectum is given by: \(\frac{2b^2}{a}\)

    CALCULATION:

    Here, we have to find the equation of the hyperbola whose foci are (± 5, 0) and the conjugate axis is of length 8.

    By comparing the foci (± 5, 0) with (± ae, 0)

    ⇒ ae = 5

    ∵ Length of the conjugate axis is given by 2b

    ⇒ 2b = 8

    ⇒ b = 4

    As we know that, eccentricity of a hyperbola is given by \(e = \frac{{\sqrt {{a^2} + {b^2}} }}{a}\)

    ⇒ a2e2 = a2 + b2

    ⇒ 25 = a2 + 16

    ⇒ a2 = 9

    So, the equation of the required hyperbola is \(\frac{{{x^2}}}{{{9}}} – \frac{{{y^2}}}{{{16}}} = 1\)

    Hence, option A is the correct answer.

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