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If the ellipse (x^2)/(a^2)+(y^2)/(b^2)=1...

If the ellipse `(x^2)/(a^2)+(y^2)/(b^2)=1` is inscribed in a rectangle whose length to breadth ratio is `2:1` , then the area of the rectangle is (a)`4.(a^2+b^2)/7` (b) `4.(a^2+b^2)/3` (c)`12.(a^2+b^2)/5` (d) `8.(a^2+b^2)/5`

A

`4*(a^(2)+b^(2))/(7)`

B

`4*(a^(2)+b^(2))/(3)`

C

`12*(a^(2)+b^(2))/(5)`

D

`8*(a^(2)+b^(2))/(5)`

Text Solution

Verified by Experts

Since mutually perpendicualar tangents can be drawn from the vartices of the reactangle , all the vertices of the rectangle should lie on the director circle `x^(2)+y^(2)=a^(2)+b^(2)`

Let breadht =2l and length =4l . then,
`l^(2)+(2l)^(2)=a^(2)+b^(2)`
or `l^(2)=(a^(2)+b^(2))/(5)`
or Area of `4lxx2l=8((a^(2)+b^(2))/(5))`
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