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प्रश्न
Draw a rough sketch of the curve and find the area of the region bounded by curve y2 = 8x and the line x =2.
उत्तर
Given equation is y2 = 8x
Comparing with y2 -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
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