#### Question

Draw a rough sketch of the curve and find the area of the region bounded by curve y^{2} = 8x and the line x =2.

#### Solution

Given equation is y^{2} = 8x

Comparing with y^{2} -4ax

we get 4a = 8

i.e a = 2

Given `y^2 = 4(2)x`

`y^2 = 8x``

`:. y = sqrt(8x)`

Also x = 2 meets `y^2 = 8x`

`:. y^2 = 16`

`:. y = +-4`

:. (2,4) and (2,-4) are their point of intersection

:. Required are `A = 2int_0^2 sqrt(8x) dx`

`=2sqrt8 int_0^2 x^("1/2") dx`

`= 4sqrt2 = [x^("3/2")/("3/2")]_0^2`

`= (8sqrt2)/3 [2^"3/2" - 0]`

`= (8sqrt2)/3 xx sqrt8`

`= (8sqrt2xx2sqrt2)/3 = 32/3` sq.units

Is there an error in this question or solution?

#### APPEARS IN

Solution Draw a Rough Sketch of the Curve and Find the Area of the Region Bounded by Curve Y2 = 8x and the Line X =2. Concept: Area of the Region Bounded by a Curve and a Line.