Question

Find the equation of the tangent to the hyperbola:

3x² – 4y² = 12 at the point (4, 3)

Answer 1

The equation of the hyperbola is 3x² – 4y² = 12, i.e., `x^2/4 - y^2/3` = 1

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

a² = 4, b² = 3

The equation of the tangent to the hyperbola

`x^2/"a"^2 - y^2/"b"^2` = 1 at the point P(x₁, y₁) on it is

`("xx"_1)/"a"^2 - ("yy"_1)/"b"^2` = 1

∴ the equation of the tangent to the given hyperbola at the point (4, 3) is

`("x"(4))/4 - ("y"(3))/3 = 1`

∴ x – y = 1