Find the equation of the tangent at t = 2 to the parabola y2 = 8x. (H

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

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

Answer»

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