Question

Find the equation of the tangent at t = 2 to the parabola y² = 8x. (Hint: use parametric form)

Answer 1

y² = 8x. 

Comparing this equation with y² = 4ax 

we get 4a = 8 ⇒ a = 2 

Now, the parametric form for y² = 4ax is x = at², y = 2at 

Here a = 2 and t = 2 

⇒ x = 2(2)² = 8 and y = 2(2) (2) = 8 

So the point is (8, 8) 

Now eqution of tangent to y² = 4 ax at (x₁, y₁) is yy₁ = 2a(x + x₁) 

Here (x₁, y₁) = (8, 8) and a = 2 

So equation of tangent is y(8) = 2(2) (x + 8) 

(ie.,) 8y = 4 (x + 8) 

(÷ by 4) ⇒ 2y = x + 8 ⇒ x – 2y + 8 = 0 

Alternative

The equation of tangent to the parabola y² = 4ax at ‘t’ is 

yt = x + at² 

Here t = 2 and a = 2 

So equation of tangent is 

(i.e.,) y(2) = x + 2(2)² 

2y = x + 8 ⇒ x – 2y + 8 = 0