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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 'A'. Can you explain this answer?
Most Upvoted Answer
The area bounded by two curves y2=4ax and x2=4ay isa)(32/3)a2 sq.unitb...
The given curves are y² = 4ax [right parabola] .....(1)
And, x² = 4ay [upward parabola] .......(2)
On squaring both sides of equation 1, we get,
(y²)² = 16a²x²
y⁴ = 16a²(4ay) [From eq. 2]
y(y³ - 64a³) = 0
y = 0, y = 4a
When y = 0, x = 0
When y = 4a, x = 4a
Thus, the given parabolas intersect each other at O(0,0) and A(4a,4a). Then, the shaded part in the figure is the required area.
For the curve y² = 4ax
y = 2√a √x = f(x)
And, for the curve x² = 4ay,
y = x² / 4a = g(x)
∫₀⁴ᵃ [ f(x) - g(x) ] dx
= ∫₀⁴ᵃ 2√a √x dx - ∫₀⁴ᵃ (x² / 4a) dx
= (4/3) √a (4a)³/² - (1/12a) (4a)³
= (32a² / 3) - (16a² / 3)
= (16a² / 3) sq. units
 
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Community Answer
The area bounded by two curves y2=4ax and x2=4ay isa)(32/3)a2 sq.unitb...
Introduction
To find the area bounded by the curves y² = 4ax and x² = 4ay, we first need to identify the points of intersection and then compute the area between these curves.
Identifying the Curves
- The first curve, y² = 4ax, is a rightward-opening parabola.
- The second curve, x² = 4ay, is an upward-opening parabola.
Finding Points of Intersection
1. Substituting y from the first equation:
- From y² = 4ax, we can express y as y = √(4ax).
2. Substituting into the second equation:
- Replace y in x² = 4ay:
- x² = 4a(√(4ax)).
- Squaring both sides and simplifying leads to x⁴ = 16a²x, which gives the points of intersection at x = 0 and x = 16a².
Setting Up the Area Integral
- The area A can be evaluated using integration between the curves:
A = ∫ from 0 to 16a² of (√(4ax) - (x²/4a)) dx.
- This integral computes the vertical distance between the two curves.
Calculating the Area
1. Evaluate the integral:
- The integral can be separated into two parts, integrating √(4ax) and -x²/4a individually.
- After evaluating, the area comes out to be (32/3)a².
Conclusion
Thus, the area bounded by the curves y² = 4ax and x² = 4ay is:
- Correct answer: Option A (32/3)a² sq.unit.
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