Question

The minimum distance between the curves x² +y² –12x+31 = 0 and y² = 4x is

(a) \(\sqrt{5}\)

(b) \(2\sqrt{5}\)

(c) \(6-\sqrt{5}\)

(d) None of these.

Answer 1

Centre (6,0) radius = \(\sqrt{36+0-31}=\sqrt{5}\)

Minimum distance obtained along the common normal.

y² = 4x

Differentiate w.r.t.x

2y\(\frac{dy}{dx}=4\)

\(\frac{dy}{dx}=\frac{2}{y}\)

slope of normal at (x₁ ,y₁) is – \(\frac{y_1}2\)

Also slope of CQ = \(\frac{y_1-0}{x_1-6}=-\frac{y_1}{2}\)

⇒ y₁ = 0 or x₁ = 4

Points are (0,0), (4,4), (4,–4)

OC = 6

QC = \(2\sqrt{5}\)

RC = \(2\sqrt{5}\)

Minimum distance \(\begin{Bmatrix}OC-R=6-\sqrt{5}\\QC-R=2\sqrt{5-\sqrt{5}=\sqrt{5}}\\RC-R=2\sqrt{5}\,-\sqrt{5}=\sqrt{5}\end{Bmatrix}\) is \(\sqrt{5}\)