Find the equation of tangents which are drawn from point (4, 10) to pa

1.	Find the equation of tangents which are drawn from point (4, 10) to parabola y2 = 8x.
Answer» From point (x₁, y₁) two tangents can be drawn at parabola whose combined equation can be find by equation SS’ = T² Given Point – (4, 0) Equation of parabola, ⇒ y² = 8x where a = 2 then S = y² – 4ax₁ ⇒ S = y² – 8x …(i) S’ = y1² – 4ax² ⇒ S’ = (10)² – 4 × 2 × 4 ⇒ S’ = 100 – 32 = 68 …(ii) T = yy1 – 2a(x + x₁) ⇒ T = y × 10 – 2 × 2 × (x + 4) = T = 10y – 4(x + 4) T² = {10y – 4(x + 4)}² ⇒ T² = 100 y² + 16(x + 4)² – 80(x + 4)y ⇒ T² = 100y² + 16x² + 256 + 128x – 80xy- 320y …(iii) Put the values from eqn. (i), (ii), (iii) in SS’ – T² (y² – 8x) (68) = 16x² + 100y² – 80xy + 128x – 320y + 256 ⇒ 68y² – 744x – 16x² – 100y² + 80xy – 128x+ 320y – 256 = 0 ⇒ – 16x² – 32y² + 80xy – 672x + 320y – 256 = 0 ⇒ x² + 2y² – 5xy + 42x – 20y + 16 = 0 This is required equation.

Find the equation of tangents which are drawn from point (4, 10) to parabola y2 = 8x.

Answer»

From point (x₁, y₁) two tangents can be drawn at parabola whose combined equation can be find by equation SS’ = T²

Given Point – (4, 0)

Equation of parabola,

⇒ y² = 8x where a = 2

then S = y² – 4ax₁

⇒ S = y² – 8x …(i)

S’ = y1² – 4ax²

⇒ S’ = (10)² – 4 × 2 × 4

⇒ S’ = 100 – 32 = 68 …(ii)

T = yy1 – 2a(x + x₁)

⇒ T = y × 10 – 2 × 2 × (x + 4)
= T = 10y – 4(x + 4)

T² = {10y – 4(x + 4)}²

⇒ T² = 100 y² + 16(x + 4)² – 80(x + 4)y

⇒ T² = 100y² + 16x² + 256 + 128x – 80xy- 320y …(iii)

Put the values from eqn. (i), (ii), (iii) in SS’ – T²

(y² – 8x) (68) = 16x² + 100y² – 80xy + 128x – 320y + 256

⇒ 68y² – 744x – 16x² – 100y²
+ 80xy – 128x+ 320y – 256 = 0

⇒ – 16x² – 32y² + 80xy – 672x + 320y – 256 = 0

⇒ x² + 2y² – 5xy + 42x – 20y + 16 = 0

This is required equation.