Find the equation of tangent to the parabola: y2 = 36x from the point

1.	Find the equation of tangent to the parabola: y2 = 36x from the point (2, 9)
Answer» Given equation of the parabola is y² = 36x. Comparing this equation with y² = 4ax, we get ⇒ 4a = 36 ⇒ a = 9 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, 9), ⇒ 9 = 2m + 9/m ⇒ 9m = 2m² + 9 ⇒ 2m² – 9m + 9 = 0 ⇒ 2m² – 6m – 3m + 9 = 0 ⇒ 2m(m – 3) – 3(m – 3) = 0 ⇒ (m – 3)(2m – 3) = 0 ⇒ m = 3 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 – 9 = 3(x – 2) and y – 9 = 3/2 (x – 2) ⇒ y – 9 = 3x – 6 and 2y – 18 = 3x – 6 ⇒ 3x – y + 3 = 0 and 3x – 2y + 12 = 0

Find the equation of tangent to the parabola: y2 = 36x from the point (2, 9)

Answer»

Given equation of the parabola is y² = 36x.

Comparing this equation with y² = 4ax, we get

⇒ 4a = 36

⇒ a = 9

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, 9),

⇒ 9 = 2m + 9/m

⇒ 9m = 2m² + 9

⇒ 2m² – 9m + 9 = 0

⇒ 2m² – 6m – 3m + 9 = 0

⇒ 2m(m – 3) – 3(m – 3) = 0

⇒ (m – 3)(2m – 3) = 0

⇒ m = 3 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 – 9 = 3(x – 2) and y – 9 = 3/2 (x – 2)

⇒ y – 9 = 3x – 6 and 2y – 18 = 3x – 6

⇒ 3x – y + 3 = 0 and 3x – 2y + 12 = 0