Statement 1- An equation of a common tangent to the parabola y2-16-3x and the ellipse 2x2-y2-4 is y-2x-2-3-Statement 2- If the line y-mx-4-3-m-m- 0 is a common tangent to the parabola y2-16-3x and the ellipse 2x2-y2-4- then m satisfies m4-2m2-24

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Solving Linear Differential Equations of First Order

Statement 1: ...

Question

Statement 1: An equation of a common tangent to the parabola $y^{2} = 16 \sqrt{3} x$ and the ellipse $2 x^{2} + y^{2} = 4$ is $y = 2 x + 2 \sqrt{3}$ .
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 $m^{4} + 2 m^{2} = 24$

Statement 1 is false, Statement 2 is true.

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Statement 1 is true, Statement 2 is true, Statement 2 is a correct expalnation for Statement 1

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Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation for Statement 1.

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Statement 1 is true , Statement 2 is false.

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Solution

The correct option is D Statement 1 is true, Statement 2 is true, Statement 2 is a correct expalnation for Statement 1
Equation of the parabola $y^{2} = 16 \sqrt{3} x$
Equation of the ellipse $2 x^{2} + y^{2} = 4$ and common tangent is $y = 2 x + 2 \sqrt{3}$
Hence, the common tangent is $y = m x + \frac{4 \sqrt{3}}{m}$
Condition of tangency in ellipse is $c^{2} = a^{2} m^{2} + b^{2}$
$\frac{48}{m^{2}} = 2 m^{2} + 4$
$\Rightarrow m^{4} + 2 m^{2} = 24$

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and the ellipse

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

is a common tangent to the parabola

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and the ellipse

2 x^{2} + y^{2} = 4

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Statement 1: An equation of a common tangent to the parabola y2=16√3x and the ellipse 2x2+y2=4 is y=2x+2√3.Statement 2: If the line y=mx+4√3m,(m≠0) is a common tangent to the parabola y2=16√3x and the ellipse 2x2+y2=4, then m satisfies m4+2m2=24