Question

Find equation of the tangent and normal to the parabola y² = 6x at the positive end of the latus rectum. 

Answer 1

Equation of parabola y² = 6x 

 4a = 6 ⇒ a = 3/2 

 Positive end of the Latus rectum is (a, 2a) = (3/2 ,3)  

Equation of tangent yy₁ = 2a(x + x₁) 

 yy₁ = 3(x + x₁)

3y=3(x+3/2)

2y – 2x – 3 = 0 is the equation of tangent 

Slope of tangent is 1 

Slope of normal is –1 

Equation of normal is

2x + 2y – 9 = 0