If the tangents drawn from the point (-6, 9) to the parabola y2 = kx a

1.	If the tangents drawn from the point (-6, 9) to the parabola y2 = kx are perpendicular to each other, find k.
Answer» Given equation of the parabola is y² = kx Comparing this equation with y² = 4ax, we get ⇒ 4a = k ⇒ a = k/4 Equation of tangent to the parabola y² = 4ax having slope m is y = mx + \(\frac {a}{m}\) Since the tangent passes through the point (-6, 9), ⇒ 9 = -6m + k/4m ⇒ 36m = -24m2 + k ⇒ 24m² + 36m – k = 0 The roots m₁ and m₂ of this quadratic equation are the slopes of the tangents. m₁ m₂ = -k/24 Since the tangents are perpendicular to each other m₁ m₂ = -1 ⇒ -k/24 = -1 ⇒ k = 24 Alternate method: We know that, tangents drawn from a point on directrix are perpendicular. (-6, 9) lies on the directrix x = -a. ⇒ -6 = -a ⇒ a = 6 Since 4a = k ⇒ k = 4(6) = 24

If the tangents drawn from the point (-6, 9) to the parabola y2 = kx are perpendicular to each other, find k.

Answer»

Given equation of the parabola is y² = kx

Comparing this equation with y² = 4ax, we get

⇒ 4a = k

⇒ a = k/4

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

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

Since the tangent passes through the point (-6, 9),

⇒ 9 = -6m + k/4m

⇒ 36m = -24m2 + k

⇒ 24m² + 36m – k = 0

The roots m₁ and m₂ of this quadratic equation are the slopes of the tangents.

m₁ m₂ = -k/24

Since the tangents are perpendicular to each other

m₁ m₂ = -1

⇒ -k/24 = -1

⇒ k = 24

Alternate method:

We know that, tangents drawn from a point on directrix are perpendicular.

(-6, 9) lies on the directrix x = -a.

⇒ -6 = -a

⇒ a = 6

Since 4a = k

⇒ k = 4(6) = 24