Question

If the point P lies on the curve 4x¢ + Sy- — 20 = 0 is farthest from the point O(0.—4) then (PQ is equal to :

Answer 1

Given : The point P lies on the curve 4x² + 5y² - 20 = 0 is farthest from the point Q(0, -4)

To find : The value of (PQ)² is equal to

solution : curve 4x² + 5y² - 20 = 0 is an ellipse.

let's convert it into standard from to write parametric equation of point P.

4x²/20 + 5y²/20 = 20/20

⇒x²/5 + y²/4 = 1

now parametric equation of point P(√5cosθ, 2sinθ)

now using distance formula,

PQ = √{(√5cosθ - 0)² + (2sinθ + 4)²}

⇒PQ² = 5cos²θ + (2sinθ + 4)²

= 5cos²θ + 4sin²θ + 16 + 16sinθ

= 5(cos²θ + sin²θ) - sin²θ + 16 + 16sinθ + 16

= 21 - sin²θ + 16 sinθ

= 85 - (sin²θ - 16sinθ + 64)

= 85 - (sinθ - 8)²

here it is clear that PQ² will be maximum when sinθ will be maximum. max value of sine is 1

so, sinθ = 1 = sin90° ⇒θ = 90°

now point P = (√5 , 0)

Therefore the value PQ² is equal to 85 - 49 = 36