JEE Advance - Mathematics (2023 - Paper 1 Online - No. 2)
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 $P: y^2=12 x$. 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 area of the quadrilateral $A_1 A_2 A_3 A_4$ is 35 square units
The area of the quadrilateral $A_1 A_2 A_3 A_4$ is 36 square units
The tangents $T_1$ and $T_2$ meet the $x$-axis at the point $(-3,0)$
The tangents $T_1$ and $T_2$ meet the $x$-axis at the point $(-6,0)$
Explanation
$$
\begin{aligned}
& E: \frac{x^2}{6}+\frac{y^2}{3}=1 \text {, Tangent : } y=m_1 x \pm \sqrt{6 m_1^2+3} \\\\
& P: y^2=12 x, \text { Tangent: } y=m_2 x+\frac{3}{m_2}
\end{aligned}
$$
For common tangent
$$ \begin{aligned} & m=m_1=m_2, \pm \sqrt{6 m_1^2+3}=\frac{3}{m_2} \\\\ & \Rightarrow m= \pm 1 \end{aligned} $$
$\Rightarrow$ Equation of common tangents $y=x+3$ and $y=-x-3$ point of contact for parabola is $\left(\frac{a}{m^2}, \frac{2 a}{m}\right)$
$$ \Rightarrow A_1 \equiv(3,6), \quad A_4(3-6) $$
Let $A_2\left(x_1, y_1\right) \Rightarrow$ tangent to $E$ is $\frac{x x_1}{6}+\frac{y y_1}{3}=1$
$A_3$ is mirror image of $A_2$ in $x$-axis $\Rightarrow A_3(-2,-1)$
Intersection point of $T_1=0$ and $T_2=0$ is $(-3,0)$
Area of quadrilateral $A_1 A_2 A_3 A_4=\frac{1}{2}(12+2) \times 5=35$ square units
For common tangent
$$ \begin{aligned} & m=m_1=m_2, \pm \sqrt{6 m_1^2+3}=\frac{3}{m_2} \\\\ & \Rightarrow m= \pm 1 \end{aligned} $$
$\Rightarrow$ Equation of common tangents $y=x+3$ and $y=-x-3$ point of contact for parabola is $\left(\frac{a}{m^2}, \frac{2 a}{m}\right)$
$$ \Rightarrow A_1 \equiv(3,6), \quad A_4(3-6) $$
Let $A_2\left(x_1, y_1\right) \Rightarrow$ tangent to $E$ is $\frac{x x_1}{6}+\frac{y y_1}{3}=1$
$A_3$ is mirror image of $A_2$ in $x$-axis $\Rightarrow A_3(-2,-1)$
Intersection point of $T_1=0$ and $T_2=0$ is $(-3,0)$
Area of quadrilateral $A_1 A_2 A_3 A_4=\frac{1}{2}(12+2) \times 5=35$ square units
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