Question

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`

Answer 1

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