If the shortest distance of the parabola y2 = 4x from the centre of the circle x2 + y2 – 4x

Engineering

Mathematics

Tangent and Normal

Question

If the shortest distance of the parabola y² = 4x from the centre of the circle x² + y² – 4x – 16y + 64 = 0 is d, then d² is equal to:

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y² = 4x ; x² + y² – 4x – 16y + 64 = 0
centre (2,8) radius = 2
∵ shortest distance will lie along common normal
∵ normal at p(t) to parabola is
∴ y + tx = 2t + t³, now it will pass through (2, 8)
⇒ 8 + 2t = 2t + t³.
t = 2 ∴ p(t) = p(t², 2t) ≡ P(4,4)
$∴ d = \sqrt{4 + 16} = \sqrt{20}$

∴ d² = 20