wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

If the ellipse x2a2+y2b2=1 is inscribed in a rectangle whose length to breadth ratio is 2:1 in such a manner that it touches the sides of rectangle, then the area of the rectangle is

A
4(a2+b27)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
4(a2+b23)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
12(a2+b25)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
8(a2+b25)
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

The correct option is D 8(a2+b25)
As given that, ellipse is inscribed in a rectangle so the length of major axis is equal to the length of rectangle and length of minor axis is equal to the breadth of rectangle.
2a2b=21a=2b (1)

Area of rectangle
=2a×2b=4ab=8b2 ( from equation (1))
Now, check
8(4b2+b25)=8b2

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
Join BYJU'S Learning Program
CrossIcon