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

Statement 2: If the line $y= mx+\frac{4\sqrt{3}}{m},(m\neq 0)$ is a common tangent to the parabola

$y^{2}= 16\sqrt{3}x$ and the ellipse $2x^{2}+y^{2}= 4,$ then m satisfies $m^{4}+2m^{2}= 24$

Option 1)
Statement 1 is false, statement 2 is true

Option 2)
Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1

Option 3)
Statement 1 is true, statement 2 is true; statement 2 is not a correct explanation for statement 1

Option 4)
Statement 1 is true, statement 2 is false

Answers (2)

best_answer

As we learnt in

Equation of tangent -

$y= mx+\frac{a}{m}$

- wherein

Tengent to $y^{2}=4ax$ is slope form.

Statement 1 :

$y^{2}= 16\sqrt{3}x,y= mx+\frac{4\sqrt{3}}{m}$

$\frac{x^{2}}{2}+\frac{y^{2}}{4}= 1,x= m_{1}y+\sqrt{4m^{2}_{1}+2}$

$\Rightarrow y= \frac{x}{m_{1}}-\sqrt{4+\frac{2}{{m_{1}}^{2}}},m= \frac{1}{m_{1}}$

$Now\: \: \left ( \frac{4\sqrt{3}}{m} \right )^{2}= \left ( -\sqrt{4+\frac{2}{{m_{1}}^{2}}} \right )^{2}$

$\Rightarrow \frac{48}{m^{2}}= 4+\frac{2}{{m_{1}}^{2}}= 4+2m^{2}\Rightarrow \frac{24}{m^{2}}= 2+m^{2}$

$\Rightarrow m^{4}+2m^{2}-24=0\cdots \cdots \cdots \cdots (i)$

$\Rightarrow \left ( m^{2}+6 \right )\left ( m^{2}-4 \right )= 0\Rightarrow m\pm 2$

Statement 2 :

If $y=mx +\frac{4\sqrt{3}}{m}$ is a common tangent to $y^{2}= 16\sqrt{3}x$ and ellipse $2x^{2}+y^{2}= 4\: \: then \: \: m\: \: satisfies m^{4}+2m^{2}-24=0$

From (i) statement 2 is a correct explanation for statement 1.

Option 1)

Statement 1 is false, statement 2 is true

This option is incorrect.

Option 2)

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

This option is correct.

Option 3)

Statement 1 is true, statement 2 is true; statement 2 is not a correct explanation for statement 1

This option is incorrect.

Option 4)

Statement 1 is true, statement 2 is false

This option is incorrect.

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