Let for two distinct values of p the lines y= | vedclass

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(B) The condition for the line $y = mx + p$ to touch the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is $p^2 = a^2m^2 + b^2$. Here $a^2 = 16, b^2

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$24$

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(B) The condition for the line $y = mx + p$ to touch the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is $p^2 = a^2m^2 + b^2$. Here $a^2 = 16, b^2 = 9, m = 1$,so $p^2 = 16(1)^2 + 9 = 25$,which gives $p = \pm 5$.The points of contact are given by $\left( \mp \frac{a^2m}{p}, \pm \frac{b^2}{p} ight)$. For $p = 5$,$A = \left( -\frac{16}{5}, \frac{9}{5} ight)$. For $p = -5$,$B = \left( \frac{16}{5}, -\frac{9}{5} ight)$.The line $y = x$ intersects the ellipse $\frac{x^2}{16} + \frac{x^2}{9} = 1$,so $x^2(\frac{9+16}{144}) = 1$,$x^2 = \frac{144}{25}$,$x = \pm \frac{12}{5}$. Thus $C = \left( \frac{12}{5}, \frac{12}{5} ight)$ and $D = \left( -\frac{12}{5}, -\frac{12}{5} ight)$.The quadrilateral $ABCD$ is a parallelogram because the lines $y = x + 5$ and $y = x - 5$ are parallel,and the line $y = x$ passes through the center $(0,0)$.The area of the quadrilateral $ABCD$ can be calculated as $2 	imes 	ext{Area}(	riangle ABC)$.Using the coordinates $A(-\frac{16}{5}, \frac{9}{5})$,$B(\frac{16}{5}, -\frac{9}{5})$,$C(\frac{12}{5}, \frac{12}{5})$,the area is $2 	imes \frac{1}{2} |x_A(y_B - y_C) + x_B(y_C - y_A) + x_C(y_A - y_B)| = |-\frac{16}{5}(-\frac{9}{5} - \frac{12}{5}) + \frac{16}{5}(\frac{12}{5} - \frac{9}{5}) + \frac{12}{5}(\frac{9}{5} - (-\frac{9}{5}))| = |-\frac{16}{5}(-\frac{21}{5}) + \frac{16}{5}(\frac{3}{5}) + \frac{12}{5}(\frac{18}{5})| = |\frac{336}{25} + \frac{48}{25} + \frac{216}{25}| = \frac{600}{25} = 24$.

Let for two distinct values of $p$ the lines $y=x+p$ touch the ellipse $E: \frac{x^2}{16} + \frac{y^2}{9} = 1$ 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

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