Common Tangents to Ellipse and Parabola Area and Intersection Point

To find the correct options, we analyze the equations of the ellipse $E$ and the parabola $P$.

The given ellipse is: $E : \frac{x^2}{6} + \frac{y^2}{3} = 1$ Here, $a^2 = 6$ and $b^2 = 3$.

The given parabola is: $P : y^2 = 12x$ Here, $4a = 12 \implies a = 3$.

Step 1: Finding the equations of the common tangents

Any line tangent to the parabola $y^2 = 12x$ with slope $m$ can be written as: $y = mx + \frac{3}{m}$

For this line to be tangent to the ellipse $\frac{x^2}{6} + \frac{y^2}{3} = 1$, it must satisfy the condition of tangency $c^2 = a^2 m^2 + b^2$: $\left(\frac{3}{m}\right)^2 = 6m^2 + 3$

$\frac{9}{m^2} = 6m^2 + 3 \implies 6m^4 + 3m^2 - 9 = 0$

$2m^4 + m^2 - 3 = 0$

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

Since $m \in \mathbb{R}$, $2m^2 + 3 \neq 0$, so we get: $m^2 = 1 \implies m = \pm 1$

Thus, the two common tangents are:

• $T_1: y = x + 3$ (for $m = 1$)
• $T_2: y = -x - 3$ (for $m = -1$)

Step 2: Finding the point of intersection on the $x$-axis

Setting $y = 0$ in the equations of $T_1$ and $T_2$:

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

Thus, both tangents $T_1$ and $T_2$ intersect the $x$-axis at the point $(-3, 0)$. Therefore, Statement (C) is TRUE and Statement (D) is FALSE.

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Step 3: Finding the points of contact

1. For $T_1: y = x + 3$ ($m = 1$):

   • Point of contact $A_1$ on the parabola $y^2 = 12x$: $A_1 = \left(\frac{a}{m^2}, \frac{2a}{m}\right) = \left(\frac{3}{1^2}, \frac{2(3)}{1}\right) = (3, 6)$
   • Point of contact $A_2$ on the ellipse $\frac{x^2}{6} + \frac{y^2}{3} = 1$: $A_2 = \left(-\frac{a^2 m}{c}, \frac{b^2}{c}\right) = \left(-\frac{6(1)}{3}, \frac{3}{3}\right) = (-2, 1)$

2. For $T_2: y = -x - 3$ ($m = -1$):

   • Point of contact $A_4$ on the parabola $y^2 = 12x$: $A_4 = \left(\frac{a}{m^2}, \frac{2a}{m}\right) = \left(\frac{3}{(-1)^2}, \frac{2(3)}{-1}\right) = (3, -6)$
   • Point of contact $A_3$ on the ellipse $\frac{x^2}{6} + \frac{y^2}{3} = 1$: $A_3 = \left(-\frac{a^2 m}{c}, \frac{b^2}{c}\right) = \left(-\frac{6(-1)}{-3}, \frac{3}{-3}\right) = (-2, -1)$

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Step 4: Area of the quadrilateral $A_1A_2A_3A_4$

The vertices of the quadrilateral are: $A_1(3, 6), \quad A_2(-2, 1), \quad A_3(-2, -1), \quad A_4(3, -6)$

Notice that $A_1A_4$ is a vertical line along $x = 3$ and $A_2A_3$ is a vertical line along $x = -2$. Thus, $A_1A_2A_3A_4$ forms a trapezium with parallel sides $A_1A_4$ and $A_2A_3$.

• Length of parallel side $A_1A_4 = 6 - (-6) = 12$
• Length of parallel side $A_2A_3 = 1 - (-1) = 2$
• Height of the trapezium $h = 3 - (-2) = 5$

The area of trapezium $A_1A_2A_3A_4$ is given by: $\text{Area} = \frac{1}{2} \times (\text{sum of parallel sides}) \times \text{height}$ $\text{Area} = \frac{1}{2} \times (12 + 2) \times 5 = \frac{1}{2} \times 14 \times 5 = 35 \text{ square units}$

Therefore, Statement (A) is TRUE and Statement (B) is FALSE.

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Conclusion

The correct statements are (A) and (C).