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

Aliter:

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