The area (in sq. units) of the smaller of the two circles that touch the parabola, y2 = 4x at the point (1, 2) and the x‑axis is:

Engineering

Mathematics

Tangent of Parabola

Question

The area (in sq. units) of the smaller of the two circles that touch the parabola, y² = 4x at the point (1, 2) and the x‑axis is:

$4 π (3 + \sqrt{2})$

$8 π (2 - \sqrt{2})$

$4 π (2 - \sqrt{2})$

$8 π (3 - 2 \sqrt{2})$

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Equation of tangent to the parabola y² = 4x at

$(1, 2) is 2 y = 4 (\frac{x + 1}{2})$

⇒ y = x + 1

Equation of normal y = –x + 3

Let centre be C (3 – r, r)

Now PC² = r²

⇒ (3 – r – 1)² + (r – 2)² = r²

⇒ 2(2 – r)² = r²

⇒ r² – 8r + 8 = 0

⇒ r = 4 ± $2 \sqrt{2}$

For r = 4 + $2 \sqrt{2}$ (3 – r < 0) Not possible

So r = 4 – $2 \sqrt{2}$

Area = πr² = π(16 + 8 – $16 \sqrt{2}$

$= 8 π (3 - 2 \sqrt{2})$