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प्रश्न
Find the equation of the tangent to the hyperbola:
`x^2/144 - y^2/25` = 1 at the point whose eccentric angle is `pi/3`
उत्तर
The equation of the hyperbola is `x^2/144 - y^2/25` = 1
Comparing with `x^2/"a"^2 - y^2/"b"^2` = 1, we get,
a2 = 144, b2 = 25
∴ a = 12, b = 5
The equation of the tangent to the hyperbola `x^2/"a"^2 - y^2/"b"^2` = 1 at the point P(θ) is
`(xsectheta)/"a" - (ytantheta)/"b"` = 1
∴ the equation of the tangent to the given hyperbola at P`(pi/3)` is
`(x sec pi/3)/12 - (y tan pi/3)/5` = 1
∴ `(2x)/12 - (sqrt(3)y)/5` = 1
∴ `x/6 - (sqrt(3)y)/5` = 1
∴ `5x - 6sqrt(3)y` = 30
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