Let for two distinct values of p the lines $$y = x + p$$ touch the ellipse E : $$\frac{x^2}{4^2} + \frac{y^2}{3^2} = 1$$ at the points A and B. Let the line $$y = x$$ intersect E at the points C and D. Then the area of the quadrilateral ABCD is equal to

The given ellipse is $$\dfrac{x^{2}}{4^{2}}+\dfrac{y^{2}}{3^{2}}=1$$, that is $$\dfrac{x^{2}}{16}+\dfrac{y^{2}}{9}=1$$.

Case 1: Find the values of $$p$$ for which the line $$y = x + p$$ is tangent to the ellipse.

Substitute $$y = x + p$$ in the ellipse:

$$\dfrac{x^{2}}{16} + \dfrac{(x+p)^{2}}{9} = 1$$

Multiply by $$144$$ (the LCM of $$16$$ and $$9$$):
$$9x^{2} + 16(x+p)^{2} = 144$$

Expand the square:
$$9x^{2} + 16\bigl(x^{2}+2px+p^{2}\bigr) - 144 = 0$$

Simplify:
$$25x^{2} + 32p\,x + 16p^{2} - 144 = 0$$ $$-(1)$$

For tangency the quadratic in $$x$$ must have discriminant $$0$$:

$$\bigl(32p\bigr)^{2} - 4 \cdot 25 \bigl(16p^{2}-144\bigr) = 0$$

$$1024p^{2} - 1600p^{2} + 14400 = 0$$

$$-576p^{2} + 14400 = 0 \;\;\Longrightarrow\;\; 576p^{2} = 14400$$

$$p^{2} = 25 \;\;\Longrightarrow\;\; p = \pm 5$$

Thus the two tangent lines are
$$y = x + 5 \quad\text{and}\quad y = x - 5$$.

Case 2: Coordinates of the points of contact $$A$$ and $$B$$.

For a tangent quadratic $$ax^{2}+bx+c=0$$, at tangency $$x = -\dfrac{b}{2a}$$. In equation $$-(1)$$ we have $$a = 25,\; b = 32p$$, so

$$x = -\dfrac{32p}{2\cdot25} = -\dfrac{16p}{25}$$.

• For $$p = 5$$:
$$x_A = -\dfrac{16(5)}{25} = -\dfrac{16}{5},\qquad y_A = x_A + 5 = -\dfrac{16}{5} + \dfrac{25}{5} = \dfrac{9}{5}$$

• For $$p = -5$$:
$$x_B = -\dfrac{16(-5)}{25} = \dfrac{16}{5},\qquad y_B = x_B - 5 = \dfrac{16}{5} - \dfrac{25}{5} = -\dfrac{9}{5}$$

Hence
$$A\!\left(-\dfrac{16}{5},\;\dfrac{9}{5}\right),\quad B\!\left(\;\dfrac{16}{5},\;-\dfrac{9}{5}\right).$$

Case 3: Intersection points $$C$$ and $$D$$ of the line $$y = x$$ with the ellipse.

Put $$y = x$$ in the ellipse:

$$\dfrac{x^{2}}{16} + \dfrac{x^{2}}{9} = 1 \;\;\Longrightarrow\;\; x^{2}\Bigl(\dfrac{1}{16}+\dfrac{1}{9}\Bigr) = 1$$

$$x^{2}\Bigl(\dfrac{25}{144}\Bigr)=1 \;\;\Longrightarrow\;\; x^{2}=\dfrac{144}{25}$$

$$x = \pm\dfrac{12}{5}\;\;,\;\; y = x$$

Therefore
$$C\!\left(\dfrac{12}{5},\;\dfrac{12}{5}\right),\quad D\!\left(-\dfrac{12}{5},\;-\dfrac{12}{5}\right).$$

Case 4: Area of quadrilateral $$ABCD$$ (vertices taken in order $$C \rightarrow A \rightarrow D \rightarrow B$$, which goes anticlockwise).

Using the Shoelace Theorem:

Coordinates:
$$C\left(\dfrac{12}{5},\;\dfrac{12}{5}\right),\; A\left(-\dfrac{16}{5},\;\dfrac{9}{5}\right),\; D\left(-\dfrac{12}{5},-\dfrac{12}{5}\right),\; B\left(\;\dfrac{16}{5},-\dfrac{9}{5}\right)$$

Compute $$\sum x_i\,y_{i+1}$$:
$$\dfrac{12}{5}\!\cdot\!\dfrac{9}{5} + \left(-\dfrac{16}{5}\right)\!\cdot\!\left(-\dfrac{12}{5}\right) + \left(-\dfrac{12}{5}\right)\!\cdot\!\left(-\dfrac{9}{5}\right) + \dfrac{16}{5}\!\cdot\!\dfrac{12}{5} = \dfrac{108+192+108+192}{25} = \dfrac{600}{25} = 24$$

Compute $$\sum y_i\,x_{i+1}$$:
$$\dfrac{12}{5}\!\cdot\!\left(-\dfrac{16}{5}\right) + \dfrac{9}{5}\!\cdot\!\left(-\dfrac{12}{5}\right) + \left(-\dfrac{12}{5}\right)\!\cdot\!\dfrac{16}{5} + \left(-\dfrac{9}{5}\right)\!\cdot\!\dfrac{12}{5} = \dfrac{-192-108-192-108}{25} = -\dfrac{600}{25} = -24$$

Area
$$=\dfrac{1}{2}\Bigl|\;\sum x_i y_{i+1} - \sum y_i x_{i+1}\Bigr| =\dfrac{1}{2}\bigl|24 - (-24)\bigr| =\dfrac{1}{2}\times48 = 24.$$

Hence the area of quadrilateral $$ABCD$$ is $$24$$.

Therefore, Option B is correct.