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प्रश्न
Find the equation of the hyperbola referred to its principal axes:
whose foci are at (±2, 0) and eccentricity `3/2`
उत्तर
Let the required equation of hyperbola be
`x^2/"a"^2 - y^2/"b"^2` = 1. ...(i)
Given, eccentricity(e) = `3/2`
Co-ordinates of foci are (±ae, 0).
Given co-ordinates of foci are (±2, 0).
∴ ae = 2
∴ `"a"(3/2)` = 2
∴ a = `4/3`
∴ a2 = `16/9`
Now, b2 = a2(e2 – 1)
∴ b2 = a2e2 – a2
= `4 - 16/9`
= `20/9`
∴ The required equation of hyperbola is
`x^2/((16/9)) - y^2/((20/9))` = 1, i.e., `(9x^2)/16 - (9y^2)/20` = 1.
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