Mathematics - Ellipse Question with Solution | TestHub

Given,

Equation of ellipse

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

Now equation of tangent with slope $m_1$ will be :

$T_1: y=m_{1}x\pm \sqrt{6m_1^2+3}$

And equation of parabola,

$P:y^2=12x$,

So, equation of tangent with slope $m_2$ will be:

$y=m_{2}x+\frac{3}{m_2}$

Now for common tangent

$m=m_1=m_2, \pm \sqrt{6m_1^2+3}=\frac{3}{m_2}$

$\Rightarrow m=\pm 1$

Hence, equation of common tangents will be,

$y=x+3$ and $y=–x-3$

Now we know that,

Point of contact for parabola is $\left(\frac{a}{m^2}, \frac{2a}{m}\right)$

$\Rightarrow A_1\equiv \left(3, 6\right), A_4\left(3-6\right)$

Now let $A_2\left(x_1, y_1\right)$

So, equation of tangent to ellipse $E$ at point $A_2\left(x_1, y_1\right)$ is given by,

 $\frac{x{x}_1}{6}+\frac{y{y}_1}{3}=1$

Now comparing above equation with $y=x+3$,

We get, $\frac{-x_1}{6}=\frac{y_1}{3}=\frac{1}{3}$

$\Rightarrow A_2\equiv \left(\frac{-6}{3},\frac{3}{3}\right)\equiv \left(-2,1\right)$

Now $A_3$ is mirror image of $A_2$ in $x$-axis 

$\Rightarrow A_3\left(-2, -1\right)$ 

Now finding the intersection point of $T_1=0$ and $T_2=0$, we get $\left(-3, 0\right)$

So, Area of quadrilateral $A_1{A}_2{A}_3{A}_4=\frac{1}{2}\left(12+2\right)\times 5=35$ square units