Let for two distinct values of p the lines y=x+p touch the ellipse E: (x^2)/(4^2)+(y^2)/(3…

Question

Let for two distinct values of p the lines $y=x+\mathrm{p}$ touch the ellipse $\mathrm{E}: \frac{x^2}{4^2}+\frac{y^2}{3^2}=1$ 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 $A B C D$ is equal to :

Accepted Answer

$E: \frac{x^2}{4^2}+\frac{y^2}{3^2}=1$

$$\begin{aligned}
& T: y=m x \pm \sqrt{16 m^2+9} \
& y=x+p \
& \Rightarrow m=1 \
& \Rightarrow p= \pm \sqrt{16+9} \
& = \pm 5
\end{aligned}$$

$T: y=x \pm 5$ will to cut the $E$ at $A\left(-\frac{16}{5}, \frac{9}{5}\right)$

$$B\left(\frac{16}{5},-\frac{9}{5}\right)$$

Also, $y=x$ will cut the $E$ at $C\left(\frac{12}{5}, \frac{12}{5}\right)$

$$D\left(-\frac{12}{5},-\frac{12}{5}\right)$$

$A B C D$ in not give in cyclic order

$\therefore$ it does not form any quadrilateral

$\therefore \quad$ No option should match

If order is not considered then

Area $=24$ sq. unit.

Solutions · 1