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

Correct answer: C. x ^ 2 + y ^ 2 - 4x + 8y + 12 = 0 y ^ 2 = 8x is the equation of the given parabola. If P is a point at a minimum distance from ' (0, - 6) ' , then it should be normal to the parabola at P . Normal to parabola y ^ 2 = 8x is y = mx - 2 · 2 · m - 2 · m ^ 3 It passes through (0, - 6) ⇒ m ^ 3 + 2m - 3 = 0 ⇒ m = 1 P(a m ^ 2 , - 2am) = P ( 2, - 4 ) Equation of circle with centre P and passes through ' C ' is (x - 2) ^ 2 + (y + 4) ^ 2 = 8 ⇒ x ^ 2 + y ^ 2 - 4x + 8y + 12 = 0

JEE Main 20163 April

Exam: JEE Main
Year: 2016
Date: 3 April
Subject: Mathematics
Chapter: Parabola

JEE Main 2016 · 3 April · Mathematics · Parabola

Let P be the point on the parabola, y ^ 2 = 8x which is at a...

Let

P

be the point on the parabola,

{y}^{2} = 8x

which is at a minimum distance from the center

C

of the circle

{x}^{2} + {\left(y + 6\right)}^{2} = 1

. Then the equation of the circle, passing through

C

and having its center at

P

Let P be the point on the parabola, y ^ 2 = 8x which is at a minimum distance from the center C of the circle x ^ 2 +

Options

A ${x}^{2} + {y}^{2} - \dfrac{x}{4} + 2y - 24 = 0$
B ${x}^{2} + {y}^{2} - 4x + 9y + 18 = 0$
C ${x}^{2} + {y}^{2} - 4x + 8y + 12 = 0$
D ${x}^{2} + {y}^{2} - x + 4y - 12 = 0$

Correct answer

C. ${x}^{2} + {y}^{2} - 4x + 8y + 12 = 0$

Step-by-step solution

{y}^{2} = 8x $is the equation of the given parabola. If$ P $is a point at a minimum distance from$ ' \left(0, - 6\right) ' $, then it should be normal to the parabola at$ P $.<img src="https://cdn-question-pool.Quantrex.app/pyq/jee_main/b2c02988-ffed-4000-8184-5ffe9df14c89-image.png" style="width: 200px; height: 236px;" />Normal to parabola$ {y}^{2} = 8x $is $ y = mx - 2 \cdot 2 \cdot m - 2 \cdot {m}^{3} $ It passes through \left(0, - 6\right)$

$⇒ {m}^{3} + 2m - 3 = 0 ⇒ m = 1$

$P\left(a{m}^{2}, - 2am\right) = P ( 2, - 4 )$

Equation of circle with centre $P$ and passes through $' C '$ is

${\left(x - 2\right)}^{2} + {\left(y + 4\right)}^{2} = 8$

$⇒ {x}^{2} + {y}^{2} - 4x + 8y + 12 = 0$

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