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प्रश्न
Answer the following:
Find the equation of the tangent to the hyperbola `x^2/25 − y^2/16` = 1 at P(30°)
उत्तर
The equation of the hyperbola is `x^2/25 − y^2/16` = 1
Comparing this with `x^2/"a"^2 - y^2/"b"^2` = 1, we get,
a2 = 25, b2 = 16
∴ a = 5, b = 4
The equation of the tangent to the hyperbola
`x^2/"a"^2 - y^2/"b"^2` = 1 at P(θ) is
`x/"a"sectheta - y/"b"tantheta` = 1
∴ the equation of the tangent to the given hyperbola at P (30°) is
`x/5sec30^circ - y/4tan30^circ` = 1
∴ `x/5(2/sqrt(3)) - y/4(1/sqrt(3))` = 1
∴ 8x – 5y = `20sqrt(3)`
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