Let T_1 and T_2 be two distinct common tangen | vedclass

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(A) The equation of a tangent to the parabola $y^2 = 12x$ is $y = mx + \frac{3}{m}$.For this line to be a tangent to the ellipse $\frac{x^2}{6} + \fra

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$A, C$

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(A) The equation of a tangent to the parabola $y^2 = 12x$ is $y = mx + \frac{3}{m}$.For this line to be a tangent to the ellipse $\frac{x^2}{6} + \frac{y^2}{3} = 1$,it must satisfy the condition $c^2 = a^2m^2 + b^2$,where $c = \frac{3}{m}$,$a^2 = 6$,and $b^2 = 3$.Substituting these values: $(\frac{3}{m})^2 = 6m^2 + 3 \implies \frac{9}{m^2} = 6m^2 + 3 \implies 3 = 2m^4 + m^2 \implies 2m^4 + m^2 - 3 = 0$.Let $u = m^2$,then $2u^2 + u - 3 = 0 \implies (2u + 3)(u - 1) = 0$. Since $u = m^2 > 0$,we have $m^2 = 1$,so $m = \pm 1$.The tangents are $y = x + 3$ and $y = -x - 3$. Both meet the $x$-axis at $(-3, 0)$. Thus,statement $(C)$ is true.For $T_1: y = x + 3$,the point of tangency on $P$ $(y^2=12x)$ is $A_1(3, 6)$ and on $E$ $(\frac{x^2}{6} + \frac{y^2}{3} = 1)$ is $A_2(-2, 1)$.For $T_2: y = -x - 3$,the point of tangency on $P$ is $A_4(3, -6)$ and on $E$ is $A_3(-2, -1)$.The quadrilateral $A_1 A_2 A_3 A_4$ is a trapezoid with parallel sides $A_1 A_4$ (length $6 - (-6) = 12$) and $A_2 A_3$ (length $1 - (-1) = 2$). The height is the distance between the lines $x = 3$ and $x = -2$,which is $3 - (-2) = 5$.Area $= \frac{1}{2} 	imes (12 + 2) 	imes 5 = \frac{1}{2} 	imes 14 	imes 5 = 35$ square units. Thus,statement $(A)$ is true.The correct statements are $(A)$ and $(C)$.

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