Question ID: #639
The area of the region, inside the ellipse $x^{2}+4y^{2}=4$ and outside the region bounded by the curves $y=|x|-1$ and $y=1-|x|$, is:
- (1) $2(\pi-1)$
- (2) $2\pi-\frac{1}{2}$
- (3) $3(\pi-1)$
- (4) $2\pi-1$
Solution:
The equation of the ellipse is $\frac{x^2}{4} + \frac{y^2}{1} = 1$.
Here $a = 2, b = 1$.
Area of ellipse $= \pi ab = \pi(2)(1) = 2\pi$.
The curves $y = |x| – 1$ and $y = 1 – |x|$ form a rhombus.
Vertices of the rhombus are found by intersections with axes:
For $y = 1 – |x|$: $(1, 0), (-1, 0), (0, 1)$.
For $y = |x| – 1$: $(1, 0), (-1, 0), (0, -1)$.
The region bounded by these curves is a square (rhombus) with diagonals of length $d_1 = 2$ (along x-axis) and $d_2 = 2$ (along y-axis).
Area of rhombus $= \frac{1}{2} d_1 d_2 = \frac{1}{2}(2)(2) = 2$.
Required Area $=$ Area of Ellipse $-$ Area of Rhombus
$= 2\pi – 2$
$= 2(\pi – 1)$
Ans. (1)
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