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 $y^2=16\sqrt{3}x$ 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-$1$.

The equation of ellipse is,
$2x^2+y^2=4$
$\therefore \frac{2x^2}{4}+\frac{y^2}{4}=1$
$\Rightarrow \frac{x^2}{2}+\frac{y^2}{4}=1$
comparing with standard form,
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
we get,
$a^2=2$ and $b^2=4$
and equation of tangent to ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is,
$y=mx\pm \sqrt{a^2{m}^2+b^2}$
$\therefore$ $y=mx\pm \sqrt{2m^2+4}...\left(1\right)$
also given that,
equation of tangent to the parabola $y^2=16\sqrt{3}x$ is,
$y=mx+\frac{4\sqrt{3}}{m}...\left(2\right)$
on comparing $\left(1\right)$ and $\left(2\right)$ we get,
$\frac{4\sqrt{3}}{m}=\sqrt{2m^2+4}$
$4\sqrt{3}=m\sqrt{2m^2+4}$
squaring on both the sides,
$48=2m^4+4m^2$
$\therefore 2m^4+4m^2-48=0$
$\Rightarrow m^4+2m^2-24=0$
$m^4+2m^2=24$ this equation satisfies statement-$2$

On solving this equation
$m^4+2m^2-24=0$
$m^4+6m^2-4m^2-24=0$
$m^2(m^2+6)-4(m^2+6)=0$
$\therefore$ we get,
$m^2+6=0$ and $m^2-4=0$
$m^2-4=0$ $\Rightarrow$$m=\pm 2$
$\therefore$ the equation of common tangent are $y=\pm 2x\pm 2\sqrt{3}$
Statement-$1$ is true, Statement-$2$ is true; Statement-$2$ is a correct explanation for statement-$1$.