Statement 1 : An equation of a common tangent to the parabola y^2=16 3 x and the ellipse 2 x^2+y^2=4 is y=2 x+2 3 . Statement 2 : If the line y=m x+ 4 3 m ,(m ≠ 0) is a common tangent to the parabola y^2=16 3 x and the

Correct answer: B. Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1 y^2=16 3 x (x^2)/(2)+(y^2)/(4)=1 y=m x+ 4 3 m is tangent to parabola which is tangent to ellipse c^2=a^2 m^2+b^2 (48)/(m^2)=2 m^2+4 m^4+2 m^2=24 m^2=4

JEE Main 2012Offline

Exam: JEE Main
Year: 2012
Mode: Offline
Subject: Mathematics
Chapter: Parabola

JEE Main 2012 · Offline · Mathematics · Parabola

Statement 1 : An equation of a common tangent to the parabola...

Statement

1

: An equation of a common tangent to the parabola

y^2=16 \sqrt{3} x

and the ellipse

2 x^2+y^2=4

y=2 x+2 \sqrt{3}

. Statement

2

: If the line

y=m x+\frac{4 \sqrt{3}}{m},(m \neq 0)

is a common tangent to the parabola

y^2=16 \sqrt{3} x

and the ellipse

2 x^2+y^2=4

, then

m

satisfies

m^4+2 m^2=24

Options

AStatement $1$ is false, statement $2$ is true
BStatement $1$ is true, statement $2$ is true; statement $2$ is a correct explanation for statement $1$
CStatement $1$ is true, statement $2$ is true; statement $2$ is not a correct explanation for statement $1$
DStatement $1$ is true, statement $2$ is false

Correct answer

B. Statement $1$ is true, statement $2$ is true; statement $2$ is a correct explanation for statement $1$

Step-by-step solution

y^2=16 \sqrt{3} x

\frac{x^2}{2}+\frac{y^2}{4}=1

y=m x+\frac{4 \sqrt{3}}{m}

is tangent to parabola which is tangent to ellipse

\Rightarrow c^2=a^2 m^2+b^2

\Rightarrow \frac{48}{m^2}=2 m^2+4 \quad \Rightarrow m^4+2 m^2=24 \quad \Rightarrow m^2=4

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