Question

Find the equation of the tangent to the hyperbola:

3x² – y² = 4 at the point `(2, 2sqrt(2))`

Answer 1

The equation of the hyperbola is 3x² – y² = 4 

i.e. `x^2/((4/3)) - y^2/4` = 1

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

a² = `4/3`, b² = 4

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

`(x(2))/((4/3)) - (y(2sqrt(2)))/4` = 1

∴ `(3x)/2 - (sqrt(2)y)/2` = 1

∴ `3x - sqrt(2)y` = 2.