Question

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 $y=mx+\frac{4\sqrt{3}}{m},(m\ne 0)$ is a common tangent to the parabola $y²=16\sqrt{3}x$ and the ellipse $2x^2+y^2=4$ then $m$ satisfies:
$m^4+2m^2=24.$

• A. Statement-$1$ is false, Statement-$2$ is true.
• B. Statement-$1$ is true, statement-$2$ is true; statement-$2$ is a correct explanation for Statement-$1$.
• C. Statement-$1$ is true, statement-$2$ is true; statement-$2$ is not a correct explanation for Statement-$1$.
• D. Statement-$1$ is true, statement-$2$ is false.

Answer 1

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

Equation of tangent to the parabola $y^2=16\sqrt{3}x$
is: $y=mx+\frac{4\sqrt{3}}{m}$
This will be tangent to the ellipse $\frac{x^2}{2}+\frac{y^2}{4}=1$
$\{\because \text{condition of tangency to ellipse is}:c^2=a^2{m}^2+b^2\}$
$\Rightarrow \frac{48}{m^2}=2m^2+4$
$\Rightarrow 48=m^2(2m^2+4)\Rightarrow 2m^4+4m^2-48=0$
$\Rightarrow m^4+2m^2-24=0$ $\Rightarrow (m^2+6)(m^2-4)=0$
$\Rightarrow m^2=4$ $\Rightarrow m=\pm 2$
equation of common tangents are $y=\pm 2x\pm 2\sqrt{3}$
$\Rightarrow$ Statement -$1$ is true.
Statement-$2$ is also true and it is the correct explanation for Statement-$1$.