Let T₁ and T₂ 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 T₁ touches P and E at the points A₁ and A₂, respectively, and the tangent T₂ touches P and E at the points A₄ and A₃, respectively. Then which of the following statements is/are true?

The correct options are:
• The area of the quadrilateral A₁A₂A₃A₄ is 35 square units
• The tangents T₁ and T₂ meet the x-axis at the point (-3, 0)

Let us analyze the given curves step-by-step and find their common tangents and points of contact.

Step 1: Equation of Common Tangents

The equation of the given parabola is:

$y^2=12x$

Here, $4a=12\Rightarrow a=3$.

Any tangent to the parabola $y^2=12x$ with slope $m$ is given by:

$y=mx+\frac{3}{m}$

Or in standard form:

$mx-y+\frac{3}{m}=0$

The equation of the given ellipse is:

$\frac{x^2}{6}+\frac{y^2}{3}=1$

Here, $a^2=6$ and $b^2=3$.

For the line $y=mx+\frac{3}{m}$ to be tangent to the ellipse, the condition of tangency $c^2=a^2{m}^2+b^2$ must be satisfied:

${\left(\frac{3}{m}\right)}^2=6m^2+3$

$\frac{9}{m^2}=6m^2+3$

Multiplying both sides by $m^2$:

$9=6m^4+3m^2\Rightarrow 2m^4+m^2-3=0$

Factoring the quadratic equation in $m^2$:

$(2m^2+3)(m^2-1)=0$

Since $m^2$ must be real and positive, we get $m^2=1\Rightarrow m=\pm 1$.

Therefore, the two common tangents are:

For $m=1$: $T_1:y=x+3$
For $m=-1$: $T_2:y=-x-3$

Step 2: Intersection of Tangents with the x-axis

Setting $y=0$ in both tangent equations:

For $T_1$: $0=x+3\Rightarrow x=-3$
For $T_2$: $0=-x-3\Rightarrow x=-3$

Thus, both tangents $T_1$ and $T_2$ meet the x-axis at the point (-3, 0).

Step 3: Finding the Points of Contact

For a parabola $y^2=4ax$, the point of contact for a tangent with slope $m$ is $\left(\frac{a}{m^2},\frac{2a}{m}\right)$.

• For $T_1$ ($m=1$), the point of contact on $P$ is $A_1=\left(\frac{3}{1^2},\frac{2\left(3\right)}{1}\right)=(3,6)$.
• For $T_2$ ($m=-1$), the point of contact on $P$ is $A_4=\left(\frac{3}{{(-1)}^2},\frac{2\left(3\right)}{-1}\right)=(3,-6)$.

For an ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, the point of contact for a tangent line $y=mx+c$ is given by $\left(-\frac{a^{2}m}{c},\frac{b^2}{c}\right)$.

• For $T_1$ ($m=1,c=3$), the point of contact on $E$ is $A_2=\left(-\frac{6\left(1\right)}{3},\frac{3}{3}\right)=(-2,1)$.
• For $T_2$ ($m=-1,c=-3$), the point of contact on $E$ is $A_3=\left(-\frac{6(-1)}{-3},\frac{3}{-3}\right)=(-2,-1)$.

Step 4: Area of Quadrilateral A₁A₂A₃A₄

The four vertices are:

$A_1=(3,6)$, $A_2=(-2,1)$, $A_3=(-2,-1)$, $A_4=(3,-6)$

Since the shape is an isosceles trapezium symmetric about the x-axis, we can calculate its area using the formula for the area of a trapezium:

$\text{Area}=\frac{1}{2}\times \left(\text{sum of parallel sides}\right)\times \left(\text{distance between parallel sides}\right)$

Parallel side 1 ($A_1{A}_4$): length = $6-(-6)=12$
Parallel side 2 ($A_2{A}_3$): length = $1-(-1)=2$
Distance between parallel sides (along x-axis): $3-(-2)=5$

Substitute these values into the area formula:

$\text{Area}=\frac{1}{2}\times (12+2)\times 5=\frac{1}{2}\times 14\times 5=35\text{square units}$

Hence, the area of the quadrilateral A₁A₂A₃A₄ is 35 square units and the tangents meet the x-axis at (-3, 0).