JEE Main 2012 — Ellipse Question with Solution

Given equation of ellipse is $$2x^2+y^2=4$$

$$\Rightarrow \frac{2x^2}{4}+\frac{y^2}{4}=1\Rightarrow \frac{x_2}{2}+\frac{y^2}{4}=1$$

Equation of tangent to the ellipse $$\frac{x^2}{2}+\frac{y^2}{4}=1$$ is

$$y=mx\pm \sqrt{2m^2+4}...\left(1\right)$$

( as equation of tangent to the ellipse $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$

is $$y=mx+c$$ where $$c=\pm \sqrt{a^2{m}^2+b^2}$$ )

Now, Equation of tangent to the parabola

$$y^2=16\sqrt{3}x$$ is $$y=mx+\frac{4\sqrt{3}}{m}...\left(2\right)$$

( as equation of tangent to the parabola

$$y^2=4ax$$ is $$y=mx+\frac{a}{m}$$ )

On comparing $$(1)$$ and $$(2),$$ we get

$$\frac{4\sqrt{3}}{m}=\pm \sqrt{2m^2+4}$$

Squaring on both the sides, we get

$$16\left(3\right)=\left(2m^2+4\right)m^2$$

$$\Rightarrow 48=m^2\left(2m^2+4\right)\Rightarrow 2m^4+4m^2-48=0$$

$$\Rightarrow m^4+2m^2-24=0\Rightarrow \left(m^2+6\right)\left(m^2-4\right)=0$$

$$\Rightarrow m^2=4$$ ( as $$m^2\ne -6$$ ) $$\Rightarrow m=\pm 2$$

$$\Rightarrow$$ Equation of common tangents are $$y=\pm 2x\pm 2\sqrt{3}$$

Thus, statement - $$1$$ is true.

Statement - $$2$$ is obviously true.