Question

Solve the following :

Find the area of the region bounded by the curve (y – 1)² = 4(x + 1) and the line y = (x – 1).

Answer 1

The equation of the curve is (y – 1)² = 4(x + 1)
This is a parabola with vertex at A(– 1, 1).
To find the points of intersection of the line y = x – 1 and the parabola.
Put y = x – 1 in the equation of the parabola, we get
(x – 1 – 1)² = 4(x + 1)
∴ x² – 4x + 4 = 4x + 4
∴ x² – 8x = 0
∴ x(x – 8) = 0
∴ x = 0, x = 8
When x = 0, y = 0 – 1 = – 1
When x = 8, y = 8 – 1 = 7
∴ the points of intersection are B(0, – 1) and C(8, 7)

To find the points where the parabola (y – 1)2 = 4(x + 1) cuts the Y-axis.
Put x = 0 in the equation of the parabola, we get
(y – 1)2 = 4(0 + 1) = 4
∴ y – 1 = ± 2
∴ y – 1 =2 or y – 1 = – 2
∴ y = 3 or y = – 1
∴ the parabola cuts the Y-axis at the points B (0, – 1) and F(0, 3).
To find the point where the line y = x – 1 cuts the X-axis. Put y = 0 in the equation of the line, we get
x – 1 = 0
∴  x = 1
∴ the line cuts the X-axis at the point G (1, 0).
Required area = area of the region BFAB +  area of the region OGDCEFO + area of the region OBGO
Now, area of the region BFAB
= area under the parabola (y – 1)² = 4(x + 1), Y-axis from y = – 1 to y = 3

= `int_(-1)^3 x*dy,  "where"  x + 1 = (y - 1)^2/(4), "i.e." x = (y - 1)^2/(4) - 1`

= `int_(-1)^3 [(y - 1)^2/(4) - 1]*dy`

= `[1/4* (y - 1)^3/(3) - y]_(-1)^3`

= `[{1/12(3 - 1)^3 - 3} - {1/12 (-1 - 1)^3 - (-1)}]`

= `(8)/(12) - 3 + (8)/(12) - 1`

= `(16)/(12) - 4`

= `(4)/(3) - 4`

= `-(8)/(3)`

Since, area cannot be negative, area of the region BFAB

= `|-(8)/(3)|`

= `(8)/(3)"sq units"`.
Area of the region OGDCEFO
= area of the region OPCEFO – area of the region GPCDG

= `int_0^8 y*dx,  "where"  (y - 1)^2`

= `4(x + 1), "i.e." y = 2sqrt(x + 1) + 1 - int_1^8y*dx,  "where"  y = x - 1`

= `int_0^8 [2sqrt(x + 1) + 1]*dx - int_1^8 (x - 1)*dx`

= `[(2*(x + 1)^(3/2))/(3/2) + x]_0^8 - [x^2/2 - x]_1^8`

= `[4/3 (9)^(3/2) + 8 - 4/3(1)^(3/2) - 0] - [(64/2 - 8) - (1/2 - 1)]`

= `(36 + 8 - 4/3) - (24 + 1/2)`

= `44 - (4)/(3) - 24 - (1)/(2)`

= `20 - (4/3 + 1/2)`

= `20 - (11)/(6)`

= `(109)/(6)"sq units"`.

Area of region OBGO = `int_0^1 y*dx,  "where"  y = x - 1`

= `int_0^1(x - 1)*dx`

= `[x^2/2 - x]_0^1`

= `(1)/(2) - 1 - 0`

= `-(1)/(2)`

Since, area cannot be negative,

area of the region = `|-(1)/(2)| = (1)/(2)"sq unit"`.

∴ required area = `(8)/(3) + (109)/(6) + (1)/(2)`

= `(16 + 109 + 3)/(6)`

= `(128)/(6)`

= `(64)/(3)"sq units"`.