Question

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

Since, the 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

Area of region OBGO =

Since, area cannot be negative, area of the region = |-1/2| = 1/2 sq units.