JEE Math Practice 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 :
- A.48
- B.20
- C.24
- D.36
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