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

Given equation of the curve is

(y − 1)² = 4(x + 1)     .......(i)

This is a parabola with vertex at A(−1, 1).

and equation of the line is y = x − 1   ......(ii)

Find the points of intersection of (y − 1)² = 4(x + 1) and y = (x − 1)

Substituting (ii) in (i), we get

(x − 1 − 1)² = 4(x + 1)

∴ x² − 4x + 4 = 4x + 4

∴ x² − 8x = 0

∴ x(x − 8) = 0

∴ x = 0 or x = 8

When x = 0, y = 0 − 1 = −1 and

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,

Substituting x = 0 in (i), 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, Substituting y = 0 in (ii), 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

= ${\displaystyle {\int }_{-1}^{3}x \text{d}y}$

= ${\displaystyle {\int }_{-1}^3[\frac{{(y-1)}^2}{4}-1]\text{d}y}$    ......[Form (i)]

= ${\displaystyle {[\frac{1}{4}\cdot \frac{{(y-1)}^3}{3}-y]}_{-1}^3}$

= ${\displaystyle [\frac{1}{12}{(3-1)}^3-3]-[\frac{1}{12}{(-1-1)}^3-(-1)]}$

= ${\displaystyle \frac{8}{13}-3+\frac{8}{12}-1}$

= ${\displaystyle \frac{4}{3}-4}$

= ${\displaystyle -\frac{8}{3}}$

= ${\displaystyle \frac{8}{3}}$    .......[∵ area cannot be negative]

Area of the region OGDCEFO = area of the region OPCEFO − area of the region GPCDG

= ${\displaystyle {\int }_0^{8}y \text{d}x-{\int }_1^{8}y \text{d}x}$

= ${\displaystyle {\int }_0^8(2\sqrt{x+1}+1)\text{d}x-{\int }_1^8(x-1) \text{d}x}$   ......[From (i) and (ii)]

= ${\displaystyle {[2\cdot \frac{{(x+1)}^{\frac{3}{2}}}{\frac{3}{2}}+x]}_0^8-{[\frac{x^2}{2}-x]}_1^8}$

= ${\displaystyle [\frac{4}{{\left(9\right)}^{\frac{3}{2}}}+8-\frac{4}{3}{\left(1\right)}^{\frac{3}{2}}-0]-[(\frac{64}{2}-8)-(\frac{1}{2}-1)]}$

= ${\displaystyle (36+8-\frac{4}{3})-(24+\frac{1}{2})}$

= ${\displaystyle 44-\frac{4}{3}-24-\frac{1}{2}}$

= ${\displaystyle 20-(\frac{4}{3}+\frac{1}{2})}$

= ${\displaystyle 20-\frac{11}{6}}$

= ${\displaystyle \frac{109}{6}}$

Area of the region OBGO = ${\displaystyle {\int }_0^{1}y \text{d}x}$

= ${\displaystyle {\int }_0^1(x-1) \text{d}x}$    ......[From (ii)]

= ${\displaystyle {[\frac{x^2}{2}-x]}_0^1}$

= ${\displaystyle \frac{1}{2}-1-0}$

= ${\displaystyle \frac{1}{2}}$    ......[∵ area cannot be negative]

∴ Required area = ${\displaystyle \frac{8}{3}+\frac{109}{6}+\frac{1}{2}}$

= ${\displaystyle \frac{16+109+3}{6}}$

= ${\displaystyle \frac{64}{3}}$ sq.units