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The area bounded by two curves y2=4ax and x2=4ay is
- a)(32/3)a2 sq.unit
- b)(16/3)sq.unit
- c)(32/3)sq.unit
- d)(16/3)a2 sq.unit
Correct answer is option 'D'. Can you explain this answer?
Verified Answer
The area bounded by two curves y2=4ax and x2=4ay isa)(32/3)a2 sq.unitb...
Given, y2 = 4ax -- (1)
x2 = 4ay ---(2)
(1) and (2) intersects
x2 = 4ay ---(2)
(1) and (2) intersects
hence
x = y2/4a (a > 0)
x = y2/4a (a > 0)
(y2/4a)2 = 4ay
y4 = 64a3y
y4 – 64a3y = 0
y[y3 – (4a)3] = 0
y = 0, 4a
When y = 0, x = 0 and when y = 4a, x = 4a.
The points of intersection of (1) and (2) are O(0, 0) and A(4a, 4a).
The area of the region between the two curves
= Area of the shaded region
= 0∫4a(y1 – y2)dx
= 0∫4a[√(4ax) – x2/4a]dx
= [2√a.(x3/2)/(3/2) – (1/4a)(x3/3)] -04a
= 4/3√a(4a)3/2 – (1/12a)(4a)3 – 0
= 32/3a2 – 16/3a2
= 16/3a2 sq. units
y4 = 64a3y
y4 – 64a3y = 0
y[y3 – (4a)3] = 0
y = 0, 4a
When y = 0, x = 0 and when y = 4a, x = 4a.
The points of intersection of (1) and (2) are O(0, 0) and A(4a, 4a).
The area of the region between the two curves
= Area of the shaded region
= 0∫4a(y1 – y2)dx
= 0∫4a[√(4ax) – x2/4a]dx
= [2√a.(x3/2)/(3/2) – (1/4a)(x3/3)] -04a
= 4/3√a(4a)3/2 – (1/12a)(4a)3 – 0
= 32/3a2 – 16/3a2
= 16/3a2 sq. units
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The area bounded by two curves y2=4ax and x2=4ay isa)(32/3)a2 sq.unitb)(16/3)sq.unitc)(32/3)sq.unitd)(16/3)a2 sq.unitCorrect answer is option 'D'. Can you explain this answer?
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