Given ellipse

$E:\ \frac{x^2}{16}+\frac{y^2}{9}=1$

We need:

• Tangents to the ellipse of the form $y=x+p$ touching at points $A,B$.
• Intersections of the line $y=x$ with the ellipse at points $C,D$.
• Then find area of quadrilateral $ABCD$.

Find tangents parallel to $y=x$

The given family is $y=x+p \quad \Longleftrightarrow \quad y-x-p=0$ or $x-y+p=0.$

For the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,$ a line $lx+my+n=0$ is tangent iff $n^2=a^2l^2+b^2m^2.$

Here, $a=4,\quad b=3,\quad l=1,\quad m=-1,\quad n=p.$

So tangent condition gives $p^2=16(1)^2+9(-1)^2=16+9=25.$

Hence $p=\pm 5.$

Therefore the two tangents are $y=x+5 \quad \text{and} \quad y=x-5.$

Find points of contact $A$ and $B$

For ellipse $\frac{x^2}{16}+\frac{y^2}{9}=1,$ tangent at point $(x_1,y_1)$ is $\frac{xx_1}{16}+\frac{yy_1}{9}=1.$

Since tangent has slope $1$, write it as $y=x+p.$ For tangent to ellipse, slope form is easier via differentiation.

Differentiate: $\frac{x^2}{16}+\frac{y^2}{9}=1$ $\frac{x}{8}+\frac{2y}{9}\frac{dy}{dx}=0$ $\frac{dy}{dx}=-\frac{9x}{16y}.$

At point of tangency, slope $=1$, so $-\frac{9x}{16y}=1$ $9x=-16y$ $y=-\frac{9x}{16}.$

Substitute into ellipse: $\frac{x^2}{16}+\frac{1}{9}\left(\frac{81x^2}{256}\right)=1$ $\frac{x^2}{16}+\frac{9x^2}{256}=1$ $\frac{16x^2+9x^2}{256}=1$ $\frac{25x^2}{256}=1$ $x^2=\frac{256}{25}$ $x=\pm \frac{16}{5}.$

Then $y=-\frac{9x}{16}$ gives $y=\mp \frac{9}{5}.$

So the two points of contact are $\left(\frac{16}{5},-\frac{9}{5}\right),\quad \left(-\frac{16}{5},\frac{9}{5}\right).$

Check with tangents:

• For $\left(\frac{16}{5},-\frac{9}{5}\right)$, $p=y-x=-\frac{9}{5}-\frac{16}{5}=-5.$
• For $\left(-\frac{16}{5},\frac{9}{5}\right)$, $p=\frac{9}{5}-\left(-\frac{16}{5}\right)=5.$

Hence we may take $A\left(-\frac{16}{5},\frac{9}{5}\right),\qquad B\left(\frac{16}{5},-\frac{9}{5}\right).$

Find points $C$ and $D$ where $y=x$ cuts the ellipse

Put $y=x$ in $\frac{x^2}{16}+\frac{y^2}{9}=1$ to get $\frac{x^2}{16}+\frac{x^2}{9}=1$ $x^2\left(\frac{1}{16}+\frac{1}{9}\right)=1$ $x^2\cdot \frac{25}{144}=1$ $x^2=\frac{144}{25}$ $x=\pm \frac{12}{5}.$

Since $y=x$, $C\left(\frac{12}{5},\frac{12}{5}\right),\qquad D\left(-\frac{12}{5},-\frac{12}{5}\right).$

Observe the shape

Slopes:

• Line through $A,B$ is $y=-\frac{9}{16}x$.
• Line through $C,D$ is $y=x$.

Also,

• Midpoint of $AB$ is $(0,0)$.
• Midpoint of $CD$ is $(0,0)$.

So diagonals $AB$ and $CD$ bisect each other. Hence $ABCD$ is a parallelogram.

Its area is $\frac{1}{2}\times (\text{product of diagonals})\times \sin\theta,$ where $\theta$ is angle between diagonals.

Lengths of diagonals

**Diagonal $AB$: **

Angle between diagonals

Slopes are $m_{AB}=-\frac{9}{16},\qquad m_{CD}=1.$

For angle between lines,

Use determinant / shoelace formula

Take vertices in order

Final answer

$\boxed{24}$

So the correct option is C.