Question

Area enclosed by the region given by
$\dfrac{ |x| }{2} \dfrac{ |y| }{3} \geqslant 1 \: and \: \dfrac{ {x}^{2} }{4} + \dfrac{ {y}^{2} }{9} \leqslant 1 \: is \:$
6x - 10
6x - 11
6x - 12
​

Answer 1

we have to find the area enclosed by the region given by, |x|/2 + |y|/3 ≥ 1 and x²/4 + y²/9 ≤ 1

solution : it can be solved easily with help of graph.

|x|/2 + |y|/3 ≥ 1

⇒3|x| + 2|y| ≥ 6 ,

if you solve it you will get four points (2, 0) , (-2, 0) , (0, 3) and (0, -3) draw the lines through the points, you will get the graph as shown in fig.

and draw the graph of ellipse , x²/4 + y²/9 ≤ 1

you will get the graph as shown in figure.

now draw both graphs in one place.

see 3rd graph, we have to find overlaps area of graphs.

so area enclosed by the region = area of ellipse - area of rhombus

= πab - 1/2 × d₁ × d₂

[ x²/4 + y²/9 ≤ 1 on comparing with x²/a² + y²/b = 1, we get a = 2 and b = 3 ]

= π × 2 × 3 - 1/2 × 4 × 6

= 6π - 12

Therefore the area enclosed by the region is 6π - 12.