Question

Find the equation of the tangent to the hyperbola:

9x² – 16y² = 144 at the point L of latus rectum in the first quadrant

Answer 1

The equation of the hyperbola is 9x² – 16y² = 144

i.e. `x^2/16 - y^2/9` = 1

Comparing with `x^2/"a"^2 - y^2/"b"^2` = 1, we get,

a² = 16, b² = 9

∴ a = 4, b = 3

Eccentricity = e = `sqrt("a"^2 + "b"^2)/"a"`

= `sqrt(16 + 9)/4`

= `5/4`

∴ ae = `4(5/4)` = 5

and `"b"^2/"a" = 9/4`

∴ the end of latus rectum in the first quadrant is

L = `("ae", "b"^2/"a") = (5, 9/4)`

The equation of the tangent to `x^2/"a"^2 - y^2/"b"^2` = 1 at the point (x₁, y₁) is `("xx"_1)/"a"^2 - ("yy"_1)/"b"^2` = 1

∴ the equation of the tangent to given hyperbola at L is

`(x(5))/16 - (y(9/4))/9` = 1

∴ `(5x)/16 - y/4` = 1 i.e., 5x – 4y = 16.