Question

If the normal at p( 18,12) to the parabola y2= 8x cuts it again at q, show that 9pq = 80 vlo

Answer 1

Answer:

9PQ = 80√10

Step-by-step explanation:

Hi,

Slope of the tangent to the parabola y² = 8x  at P(x, y) will be

2y(dy/dx) = 8

⇒ dy/dx = 4/y.

Slope of tangent at P (18, 12) is 4/12 = 1/3.

Slppe of normal at P(18, 12) will be -1/Slope of tangent

⇒ Slope of normal = -3.

Equation of the normal at P(18, 12) is given by

y-12/x-18 = -3

⇒y - 12 = -3x + 54

⇒3x + y = 66

Points of intersection of normal with parabola will be

y² = 8x

⇒ y² = 8(66 - y)/3

⇒3y² = 528 - 8y

⇒3y² + 8y - 528 = 0

⇒3y² + 44y - 36y - 528 = 0

⇒y(3y + 44) - 12 (3y + 44) = 0

⇒(y - 12)(3y + 44) = 0

=> y = 12 or y = -44/3

So, if y = 12, x = 18 this is P(18, 12)

if y = -44/3, then x = 242/9, this is Q(242/9, -44/3)

Distance between P and Q, PQ = √( 18 - 242/9)² + (12 + 44/3)²

= √1/9*[ 6400/9 + 6400] = (80√10)/9

⇒9PQ = 80√10

Hope, it helped !