Let for two distinct values of p the lines y=x+p touch the ellipse E:x^24^2+y^23^2=1 at th…

Question

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 :

Accepted Answer

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

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

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

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

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

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

$ABCD$ in not give in cyclic order

$	herefore $ it does not form any quadrilateral

$	herefore \,$ No option should match

If order is not considered then

Area $=24$ sq. unit.

Solutions · 1