Find the equation of tangent to the parabola: y2 = 12x from the point

1.	Find the equation of tangent to the parabola: y2 = 12x from the point (2, 5)
Answer» Given equation of the parabola is y² = 12x. Comparing this equation with y² = 4ax, we get ⇒ 4a = 12 ⇒ a = 3 Equation of tangent to the parabola y² = 4ax having slope m is y = mx + \(\frac {a}{m}\) Since the tangent passes through the point (2, 5) ⇒ 5 = 2m + 3/m ⇒ 5m = 2m² + 3 ⇒ 2m² – 5m + 3 = 0 ⇒ 2m² – 2m – 3m + 3 = 0 ⇒ 2m(m – 1) – 3(m – 1) = 0 ⇒ (m- 1)(2m – 3) = 0 ⇒ m = 1 or m = 3/2 These are the slopes of the required tangents. By slope point form, y – y₁ = m(x – x₁), the equations of the tangents are ⇒ y – 5 = 1(x – 2) and y – 5 = 3/2 (x – 2) ⇒ y – 5 = x – 2 and 2y – 10 = 3x – 6 ⇒ x – y + 3 = 0 and 3x – 2y + 4 = 0

Find the equation of tangent to the parabola: y2 = 12x from the point (2, 5)

Answer»

Given equation of the parabola is y² = 12x. Comparing this equation with y² = 4ax, we get

⇒ 4a = 12

⇒ a = 3

Equation of tangent to the parabola y² = 4ax having slope m is

y = mx + \(\frac {a}{m}\)

Since the tangent passes through the point (2, 5)

⇒ 5 = 2m + 3/m

⇒ 5m = 2m² + 3

⇒ 2m² – 5m + 3 = 0

⇒ 2m² – 2m – 3m + 3 = 0

⇒ 2m(m – 1) – 3(m – 1) = 0

⇒ (m- 1)(2m – 3) = 0

⇒ m = 1 or m = 3/2

These are the slopes of the required tangents.

By slope point form, y – y₁ = m(x – x₁), the equations of the tangents are

⇒ y – 5 = 1(x – 2) and y – 5 = 3/2 (x – 2)

⇒ y – 5 = x – 2 and 2y – 10 = 3x – 6

⇒ x – y + 3 = 0 and 3x – 2y + 4 = 0