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The area bounded by the parabola y2=4ax and the straight line y=2ax is
- a)a 2 3
- b)1 3a 2
- c)1 3a
- d)2 3a
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
The area bounded by the parabola y2=4ax and the straight line y=2ax is...
On solving y 2 = 4ax and y = 2ax, we get
x = 0 or 1 a
and y = 0 or 2
Most Upvoted Answer
The area bounded by the parabola y2=4ax and the straight line y=2ax is...
To find the area bounded by the parabola y^2 = 4ax and the straight line y = 2ax, we need to determine the points of intersection between the parabola and the line. Then, we can find the area using integration.
Finding the Points of Intersection:
1. Substitute y = 2ax into the equation of the parabola:
(2ax)^2 = 4ax
4a^2x^2 = 4ax
4a^2x^2 - 4ax = 0
4ax(ax - 1) = 0
2. Set each factor equal to zero:
4ax = 0 --> x = 0
ax - 1 = 0 --> x = 1/a
Therefore, the points of intersection are (0, 0) and (1/a, 2/a).
Finding the Area:
1. Determine the limits of integration. Since the x-coordinate of the points of intersection are 0 and 1/a, the limits of integration are 0 and 1/a.
2. Set up the integral for the area:
Area = ∫[0 to 1/a] (2ax - √(4ax)) dx
3. Integrate the expression:
Area = ∫[0 to 1/a] (2ax - 2√ax) dx
= 2∫[0 to 1/a] (ax - √ax) dx
4. Simplify and evaluate the integral:
Area = 2(a/2)x^2 - 2(2/3)(ax)^(3/2)] [0 to 1/a]
= a(1/a)^2 - 2(2/3)(a/2)(1/a)^(3/2) - 0
= 1 - 2/3(1/a)^(1/2)
= 1 - 2/3a^(1/2)
= 1 - 2/3√a
The area bounded by the parabola y^2 = 4ax and the straight line y = 2ax is 1 - 2/3√a, which matches the given answer option 'C'.
Finding the Points of Intersection:
1. Substitute y = 2ax into the equation of the parabola:
(2ax)^2 = 4ax
4a^2x^2 = 4ax
4a^2x^2 - 4ax = 0
4ax(ax - 1) = 0
2. Set each factor equal to zero:
4ax = 0 --> x = 0
ax - 1 = 0 --> x = 1/a
Therefore, the points of intersection are (0, 0) and (1/a, 2/a).
Finding the Area:
1. Determine the limits of integration. Since the x-coordinate of the points of intersection are 0 and 1/a, the limits of integration are 0 and 1/a.
2. Set up the integral for the area:
Area = ∫[0 to 1/a] (2ax - √(4ax)) dx
3. Integrate the expression:
Area = ∫[0 to 1/a] (2ax - 2√ax) dx
= 2∫[0 to 1/a] (ax - √ax) dx
4. Simplify and evaluate the integral:
Area = 2(a/2)x^2 - 2(2/3)(ax)^(3/2)] [0 to 1/a]
= a(1/a)^2 - 2(2/3)(a/2)(1/a)^(3/2) - 0
= 1 - 2/3(1/a)^(1/2)
= 1 - 2/3a^(1/2)
= 1 - 2/3√a
The area bounded by the parabola y^2 = 4ax and the straight line y = 2ax is 1 - 2/3√a, which matches the given answer option 'C'.
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The area bounded by the parabola y2=4ax and the straight line y=2ax isa)a 2 3b)1 3a 2c)1 3ad)2 3aCorrect answer is option 'C'. Can you explain this answer?
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