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JEE Main 20199 JanuaryMorning Shift

Exam: JEE Main
Year: 2019
Date: 9 January
Shift: Morning Shift
Subject: Mathematics
Chapter: Parabola

JEE Main 2019 · 9 January · Morning Shift · Mathematics · Parabola

Equation of a common tangent to the circle, x ^ 2 + y ^ 2 - 6x = 0...

Equation of a common tangent to the circle,

{x}^{2} + {y}^{2} - 6x = 0

and the parabola,

{y}^{2} = 4x

is:

Options

A $2\sqrt{3}y = - x - 12$
B $\sqrt{3}y = x + 3$
C $\sqrt{3}y = 3x + 1$
D $2\sqrt{3}y = 12x + 1$

Correct answer

B. $\sqrt{3}y = x + 3$

Step-by-step solution

Let tangent to parabola

{y}^{2} = 4x

is

y = mx + \dfrac{1}{m} . . . \left(i\right)

If equation

( i )

is tangent to given circle whose centre is

( 3, 0 ) and radius is

3

then length of perpendicular from centre of circle to equation

( i )

is equal to radius of circle.<br /><br />Hence, \left|\dfrac{3m + \dfrac{1}{m}}{\sqrt{{m}^{2} + 1}}\right| = 3

⇒ \left|3{m}^{2} + 1\right| = 3m\sqrt{1 + {m}^{2}}

⇒ 9{m}^{4} + 6{m}^{2} + 1 = 9{m}^{2} + 9{m}^{4}

⇒ m = \pm \dfrac{1}{\sqrt{3}}

∴

common tangents are

y = \dfrac{x}{\sqrt{3}} + \sqrt{3}

or

y = - \dfrac{x}{\sqrt{3}} - \sqrt{3}

⇒ \sqrt{3}y = x + 3or\sqrt{3}y = - x - 3

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