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Question
Find the area of the region lying between the parabolas y2 = 4ax and x2 = 4ay.
Solution
The equations of the parabolas are:
`y^2=4ax` ......(i)
`x^2=4ay` ......(ii)
`[x^2/(4a)]^2`= 4ax ......[by (ii)]
⇒ `x^4=64a^3x`
⇒ `x[x^3-(4a)^3]=0`
or x = 0 and x = 4a
∴ y = 0 and y = 4a
Points of intersection of curves are O(0, 0), P(4a, 4a)
∴ Required area = `int_0^(4a) [y_1 - y_2]dx`
`=int_0^(4a)sqrt(4ax) dx-int_0^(4a)x^2/(4a)dx`
`=sqrt(4a).2/3[x^(3/2)]_0^(4a)-1/(4a) . 1/3[x^3]_0^(4a)`
`=(4sqrta)/3xx4asqrt(4a)-1/(12a)xx64a^3`
`=32/3 a^2-16/3a^2`
`=16/3a^2` sq.units
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