Let $$T_1$$ and $$T_2$$ be two distinct common tangents to the ellipse $$E: \frac{x^2}{6} + \frac{y^2}{3} = 1$$ and the parabola $$P: y^2 = 12x$$. Suppose that the tangent $$T_1$$ touches P and E at the points $$A_1$$ and $$A_2$$, respectively and the tangent $$T_2$$ touches P and E at the points $$A_4$$ and $$A_3$$, respectively. Then which of the following statements is(are) true?

The two given curves are
ellipse $$E:\dfrac{x^{2}}{6}+\dfrac{y^{2}}{3}=1$$ and parabola $$P:y^{2}=12x$$.

Write the parabola in the standard form $$y^{2}=4ax$$.
Since $$12x=4ax$$, we have $$a=3$$.

Common tangent in slope form
  • For $$P:y^{2}=4ax$$ the tangent having slope $$m$$ is $$y=mx+\dfrac{a}{m}=mx+\dfrac{3}{m}\quad -(1)$$
  • For the ellipse $$\dfrac{x^{2}}{6}+\dfrac{y^{2}}{3}=1$$ the tangent with the same slope $$m$$ is $$y=mx\pm\sqrt{6m^{2}+3}\quad -(2)$$

Because the line must be tangent to both curves, the two intercepts in (1) and (2) must coincide:
$$\dfrac{3}{m}=\pm\sqrt{6m^{2}+3}$$ Square both sides:

$$\left(\dfrac{3}{m}\right)^{2}=6m^{2}+3 \;\Longrightarrow\;9=6m^{4}+3m^{2} \;\Longrightarrow\;6m^{4}+3m^{2}-9=0$$ Divide by $$3$$, put $$n=m^{2}\ge 0$$:

$$2n^{2}+n-3=0 \;\Longrightarrow\;n=\dfrac{-1\pm5}{4}$$ The non-negative root is $$n=1\;\Rightarrow\;m^{2}=1$$, hence $$m=1\quad\text{or}\quad m=-1$$.

Equations of the two common tangents
For $$m=1$$: from (1) $$y=x+3$$   (call this $$T_{1}$$)
For $$m=-1$$: from (1) $$y=-x-3$$ (call this $$T_{2}$$)

Intersection with the x-axis
Set $$y=0$$:
For $$T_{1}$$: $$0=x+3\;\Rightarrow\;(-3,0)$$
For $$T_{2}$$: $$0=-x-3\;\Rightarrow\;(-3,0)$$
Thus both tangents meet the x-axis at the single point $$(-3,0)$$.

Points of contact on the parabola
For $$y^{2}=4ax$$ the tangent $$y=mx+\dfrac{a}{m}$$ touches the curve at $$\left(\dfrac{a}{m^{2}},\dfrac{2a}{m}\right)$$.

• For $$m=1$$: $$A_{1}\bigl(\,3,\,6\bigr)$$
• For $$m=-1$$: $$A_{4}\bigl(\,3,\,-6\bigr)$$

Points of contact on the ellipse
Substitute each tangent in the ellipse and use the fact that a tangent cuts the curve at a repeated root.

For $$T_{1}:y=x+3$$:
$$\dfrac{x^{2}}{6}+\dfrac{(x+3)^{2}}{3}=1 \;\Longrightarrow\;(x+2)^{2}=0 \;\Rightarrow\;x=-2,\;y=1$$ Hence $$A_{2}(-2,1)$$.

For $$T_{2}:y=-x-3$$:
$$\dfrac{x^{2}}{6}+\dfrac{(-x-3)^{2}}{3}=1 \;\Longrightarrow\;(x+2)^{2}=0 \;\Rightarrow\;x=-2,\;y=-1$$ Hence $$A_{3}(-2,-1)$$.

Therefore the four vertices are $$A_{1}(3,6),\;A_{2}(-2,1),\;A_{3}(-2,-1),\;A_{4}(3,-6).$$

Area of quadrilateral $$A_{1}A_{2}A_{3}A_{4}$$ (shoelace formula)

$$$ \begin{array}{r|r} x & y \\ \hline 3 & 6 \\ -2 & 1 \\ -2 & -1 \\ 3 & -6 \\ \end{array} $$$ Sum_{1}=3$$\cdot$$1+(-2)(-1)+(-2)(-6)+3$$\cdot$$6=35\\ Sum_{2}=6(-2)+1(-2)+(-1)3+(-6)3=-35 \]

$$\text{Area}=\dfrac{1}{2}\,\bigl|\,\text{Sum}_{1}-\text{Sum}_{2}\bigr| =\dfrac{1}{2}\,(35-(-35))=\dfrac{70}{2}=35\text{ square units}.$$

Verification of the statements
A. Area is $$35$$ - TRUE.
B. Area is $$36$$ - FALSE.
C. Both tangents meet the x-axis at $$(-3,0)$$ - TRUE.
D. Meeting point $$(-6,0)$$ - FALSE.

Hence the correct options are:
Option A (area $$35$$ square units) and Option C (intersection point $$(-3,0)$$).