Let P be the point on the parabola, y2 = 8x which is at a minimum distance from the centre C of the circle, x2 + (y + 6)2 = 1. Then the equation of the circle, passing through C and having its centre at P is

Engineering

Mathematics

Normal to Parabola

Question

Let P be the point on the parabola, y² = 8x which is at a minimum distance from the centre C of the circle, x² + (y + 6)² = 1. Then the equation of the circle, passing through C and having its centre at P is

x² + y² – 4x + 8y + 12 = 0

x² + y² – 4x + 9y + 18 = 0

x² + y² – x + 4y – 12 = 0

x² + y² – + 2y – 24 = 0

LiveFree JEE Main Previous Year Online Papers Live Free JEE Advance Previous Year Online Papers

College PredictorLive

Know your College Admission Chances Based on your Rank/Percentile, Category and Home State.

Get your JEE Main Personalised Report with Top Predicted Colleges in JoSA

Normal to the parabola y² = 8x

y = mx – 2am – am³

which passes through (0, –6)

– 6 = 0 – 2am – am³

– 6 = – 4m – 2m³

m³ + 2m – 3 = 0

m = 1

$∴ P \in$ (2, – 4)

Radius of the circle, r = CP = $\sqrt{4 + 4}$

$= \sqrt{8} = 2 \sqrt{2} .$

Equation of the required circle is

(x – 2)² + (y + 4)² = 8

x² + y² – 4x + 8y + 12 = 0.