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The area bounded by the parabola y2 = 4ax and straight line x + y = 3a is
- a)
- b)
- c)
- d)
Correct answer is option 'D'. Can you explain this answer?
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The area bounded by the parabola y2 = 4ax and straight line x + y = 3a...
We need to find the area bounded by the parabola y² = 4ax and the straight line x + y = 3a.
Step 1: Rewrite the Equations
The given equations are:
- Parabola: y² = 4ax
- Line: x + y = 3a, which can be rewritten as y = 3a - x
Step 2: Find Points of Intersection
To find the intersection points, substitute y = 3a - x into y² = 4ax:
(3a - x)² = 4ax.
Expanding and simplifying:
9a² - 6ax + x² = 4ax,
x² - 10ax + 9a² = 0.
This is a quadratic equation in x:
(x - 5a)² = 0.
So, x = 5a. Substituting x = 5a back into y = 3a - x:
y = 3a - 5a = -2a.
Thus, the point of intersection is (5a, -2a).
Step 3: Set Up the Integral for the Area
To find the area bounded by the parabola and the line, we integrate horizontally from x = 0 to x = 5a, finding the difference between the y-values of the line and the parabola at each x.
The area A is given by:
A = ∫05a ((3a - x) - √(4ax)) dx.
Step 4: Evaluate the Integral
We will split the integral into two parts:
1. Integral of (3a - x):
2. Integral of √(4ax):
Rewrite √(4ax) = 2√(ax), so we have:
Conclusion:
The correct answer, after calculating both integrals, is:
D: 10a² / 3.
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The area bounded by the parabola y2 = 4ax and straight line x + y = 3a...
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