Question

The area of region $x^2+2y-1=0$, $y^2+4x-4=0$ and $y^2-4x-4=0$, in the upper half plane is ____. [ JEE MAINS 2021, July 22nd Second Shift }

Answer 1

$\large\underline{\sf{Solution-}}$

Given curves are

$\red{\rm :\longmapsto\: {x}^{2} + 2y - 1 = 0 - - - (1)}$

$\red{\rm :\longmapsto\: {y}^{2} + 4x - 4 = 0 - - - (2)}$

$\red{\rm :\longmapsto\: {y}^{2} - 4x - 4 = 0 - - - (3)}$

Now,

Consider,

$\red{\rm :\longmapsto\: {x}^{2} + 2y - 1 = 0 }$

can be rewritten as

$\red{\rm :\longmapsto\: {x}^{2}= - 2y + 1}$

$\red{\rm :\longmapsto\: {x}^{2}= - 2\bigg(y - \dfrac{1}{2}\bigg) }$

represents the downward parabola whose vertex is (0, 0.5) and intersects the x - axis at (- 1, 0) and (1, 0)

[ See the attachment graph ( green curve ) ]

Now, Consider

$\red{\rm :\longmapsto\: {y}^{2} + 4x - 4 = 0}$

can be rewritten as

$\red{\rm :\longmapsto\: {y}^{2} = - 4x + 4}$

$\red{\rm :\longmapsto\: {y}^{2} = - 4(x - 1)}$

represents the left handed parabola whose vertex is (1, 0) and intersects the y - axis at ( 0, 2 ) and ( 0, - 2 ).

[ See the attachment graph ( purple curve ) ].

Now, Consider,

$\red{\rm :\longmapsto\: {y}^{2} - 4x - 4 = 0}$

can be rewritten as

$\red{\rm :\longmapsto\: {y}^{2} = 4x + 4}$

$\red{\rm :\longmapsto\: {y}^{2} = 4(x + 1)}$

represents the right handed parabola whose vertex is (- 1, 0) and intersects the y - axis at ( 0, 2 ) and ( 0, - 2 ).

[ See the attachment graph ( black curve ) ]

So, According to statement, we have to find the area bounded by the curve in the upper half plane.

Let we assume that,

From curve (1),

$\red{\rm :\longmapsto\: {x}^{2} + 2y - 1 = 0 }$

$\red{\rm :\longmapsto\: y_1= \dfrac{1}{2} ( - {x}^{2} + 1)}$

From Curve :- 2

$\red{\rm :\longmapsto\: {y}^{2} = - 4(x - 1)}$

$\red{\rm :\longmapsto\: {y}^{2} = 4(1 - x)}$

$\red{\rm :\longmapsto\:y_2 = 2 \sqrt{1 - x}}$

From Curve :- 3

$\red{\rm :\longmapsto\: {y}^{2} = 4(x + 1)}$

$\red{\rm :\longmapsto\:y_3 = 2 \sqrt{1 + x}}$

Hence, Required area is

$\rm \: = \: 2 \: \displaystyle\int_{-1}^0\rm (y_3 - y_1) \: dx$

$\rm \: = \: 2 \bigg[\: \displaystyle\int_{-1}^0\sf 2 \sqrt{x + 1} \: dx \: - \frac{1}{2} \displaystyle\int_{-1}^0\sf ( - {x}^{2} + 1) \: dx\bigg]$

$\rm \: = \: 4 \times \dfrac{2}{3}\bigg |{\bigg(x\bigg) }^{ \dfrac{3}{2} }\bigg |_{-1}^0 - \bigg | - \dfrac{ {x}^{3} }{3} + x \bigg|_{-1}^0$

$\rm \: = \: \dfrac{8}{3}\bigg[0 - ( - 1)\bigg] - \bigg[ - \dfrac{1}{3} + 1\bigg]$

$\rm \: = \: \dfrac{8}{3} - \dfrac{2}{3}$

$\rm \: = \: \dfrac{6}{3}$

$\rm \: = \: 2 \: square \: units$