Question

Find the equations of the tangent and normal to hyperbola 12x² – 9y² = 108 at θ = `pi/3`. (Hint: use parametric form)

Answer 1

(i) Equation of the tangent to hyperbola be

`(xsectheta)/"a" - (ytantheta)/"b"` = 1

Given 12x² – 9y² = 108

`(12x^2)/108 - (9y^2)/108` = 1

`x^2/9 - y^2/12` = 1

a² = 9, b² = 12

a = 3, b = `2sqrt(3)`, θ = `pi/3`

∴ `(x sec  pi/2)/3 - (y tan  pi/3)/(2sqrt(3))` = 1

`(x(2))/3 - (ysqrt(3))/(2sqrt(3))` = 1

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

`(4x - 3y)/6` = 1

 ⇒ 4x – 3y = 6

⇒ 4x – 3y – 6 = 0

(ii) Equation of the normal to hyperbola be

`("a"x)/sec theta + ("b"y)/tan theta = "a"^2 + "b"^2`

`(3x)/(sec  pi/3) + (2sqrt(3)y)/(tan  pi/3)` = 9 + 12

`(3x)/2 + (2sqrt(3)y)/sqrt(3)` = 21

`(3x + 4y)/2` = 21

⇒ 3x + 4y = 42

⇒  3x + 4y – 42 = 0