Ellipse

Let $\left(x_1, y_1\right)$ be the point of intersection of tangents to the ellipse, where it is intersected by the line $l x+m y+n=0$. Then, the equation of the chord of contact of tangents is

$ \frac{x x_1}{a^2}+\frac{y y_1}{b^2}=1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) $

Clearly, $l x+m y+n=0$ and (i) represents the same line.

$ \begin{array}{ll} \therefore & \frac{x_1}{a^2 l}=\frac{y_1}{b^2 m}=\frac{1}{-n} \\ \Rightarrow & x_1=\frac{-a^2 l}{n}, y_1=\frac{-b^2 m}{n} \end{array} $

Thus, the required point is $\left(\frac{-a^2 l}{n}, \frac{-b^2 m}{n}\right)$

Let the rectangle have vertices at $( \pm x, \pm y)$ on the ellipse. From the ellipse equation

$ x=a \cos \theta, y=b \sin \theta $

The area of the rectangle is

$ A=4 x y=4(a \cos \theta \cdot b \sin \theta) $

Using the identity $\sin 2 \theta=2 \sin \theta \cos \theta$, we rewrite $A=2 a b \sin 2 \theta$

To maximize $A$, we maximize $\sin 2 \theta$, which reaches its maximum value of 1 when $2 \theta=\frac{\pi}{2}$.

Thus, the maximum area is

$ A_{\max }=2 a b $

Given equation of ellipse

$16{x^2} + 11{y^2} = 256 \Rightarrow {{{x^2}} \over {16}} + {{11{y^2}} \over {256}} = 1$

Equation of tangent of ellipse at $\left( {4\cos \phi ,{{16} \over {\sqrt {11} }}\sin \phi } \right)$ is

${{4x\cos \phi } \over {16}} + {{11 \times 16y} \over {256\sqrt {11} }}\sin \phi = 1$

$ \Rightarrow {{x\cos \phi } \over 4} + {{\sqrt {11} } \over {16}}\sin \phi y = 1$

is also a tangent of circle ${x^2} + {y^2} - 2x = 15$

$ \Rightarrow {(x - 1)^2} + {y^2} = 16$

radius of circle = 4 and centre = (1, 0)

Now, ${{x\cos \phi } \over 4} + {{\sqrt {11} } \over {16}}\sin \phi y = 1$ is tangent of circle.

$\therefore$ $4 = \left| {{{{{\cos \phi } \over 4} - 1} \over {\sqrt {{{{{\cos }^2}\phi } \over {16}} + {{11} \over {256}}{{\sin }^2}\phi } }}} \right| \Rightarrow \phi = \pm {\pi \over 3}$

$\therefore$ $\phi = {\pi \over 3}$