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If area bounded by the curve y^(2)=4ax a...

If area bounded by the curve `y^(2)=4ax` and `y=mx` is `a^(2)//3` , then the value of m, is

A

1

B

2

C

3

D

`sqrt3`

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The correct Answer is:
To solve the problem, we need to find the value of \( m \) such that the area bounded by the curve \( y^2 = 4ax \) and the line \( y = mx \) is equal to \( \frac{a^2}{3} \). ### Step-by-Step Solution: 1. **Identify the curves**: The first curve is a parabola given by \( y^2 = 4ax \) and the second curve is a straight line given by \( y = mx \). 2. **Find points of intersection**: To find the points of intersection of the curves, we substitute \( y = mx \) into the equation of the parabola: \[ (mx)^2 = 4ax \] This simplifies to: \[ m^2x^2 - 4ax = 0 \] Factoring out \( x \): \[ x(m^2x - 4a) = 0 \] This gives us \( x = 0 \) (the origin) and \( x = \frac{4a}{m^2} \). 3. **Set up the area integral**: The area \( A \) bounded by the curves from \( x = 0 \) to \( x = \frac{4a}{m^2} \) can be calculated using the integral: \[ A = \int_0^{\frac{4a}{m^2}} \left( \sqrt{4ax} - mx \right) \, dx \] 4. **Calculate the integral**: We first compute the integral: \[ A = \int_0^{\frac{4a}{m^2}} \sqrt{4ax} \, dx - \int_0^{\frac{4a}{m^2}} mx \, dx \] The first integral: \[ \int \sqrt{4ax} \, dx = \frac{2}{3} (4a)^{1/2} x^{3/2} = \frac{2\sqrt{4a}}{3} x^{3/2} \] Evaluating from \( 0 \) to \( \frac{4a}{m^2} \): \[ = \frac{2\sqrt{4a}}{3} \left( \frac{4a}{m^2} \right)^{3/2} = \frac{2\sqrt{4a}}{3} \cdot \frac{8a^{3/2}}{m^3} = \frac{16a^2}{3m^3} \] The second integral: \[ \int mx \, dx = \frac{m}{2} x^2 \] Evaluating from \( 0 \) to \( \frac{4a}{m^2} \): \[ = \frac{m}{2} \left( \frac{4a}{m^2} \right)^2 = \frac{m}{2} \cdot \frac{16a^2}{m^4} = \frac{8a^2}{m^3} \] 5. **Combine the results**: The total area \( A \) is: \[ A = \frac{16a^2}{3m^3} - \frac{8a^2}{m^3} = \frac{16a^2 - 24a^2}{3m^3} = \frac{8a^2}{3m^3} \] 6. **Set the area equal to \( \frac{a^2}{3} \)**: We set the area equal to \( \frac{a^2}{3} \): \[ \frac{8a^2}{3m^3} = \frac{a^2}{3} \] Dividing both sides by \( \frac{a^2}{3} \) (assuming \( a \neq 0 \)): \[ 8 = m^3 \] Thus, we find: \[ m = 2 \] ### Final Answer: The value of \( m \) is \( 2 \).
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OBJECTIVE RD SHARMA ENGLISH-AREAS OF BOUNDED REGIONS-Chapter Test
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  2. If A(1) is the area enclosed by the curve xy=1, x-axis and the ordinat...

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  3. If area bounded by the curve y^(2)=4ax and y=mx is a^(2)//3 , then the...

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  4. The value of a for which the area between the curves y^(2) = 4ax and x...

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  5. If the area bounded by the curve y=f(x), x-axis and the ordinates x=1 ...

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  6. The area bounded by the curve y = sin2x, axis and y=1, is

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  7. The area between the curve x=-2y^(2)and x=1-3y^(2), is

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  8. The area between the curves y=cosx, x-axis and the line y=x+1, is

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  9. If the area bounded by the curve y=x^2+1 and the tangents to it drawn ...

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  10. The positive value of the parmeter 'a' for which the area of the figur...

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  11. The area in square units bounded by the curves y=x^(3),y=x^(2) and the...

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  12. The area bounded by the curve y^(2)=x and the ordinate x=36 is divided...

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  13. The area contained between the x-axis and one area of the curve y=cos ...

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  14. The area of the figure bounded by |y|=1-x^(2) is in square units,

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  15. The area of the figure bounded by y=e^(x-1),y=0,x=0 and x=2 is

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  16. The area of the region on place bounded by max (|x|,|y|) le 1/2 is

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  17. The area of the closed figure bounded by y=(x^(2))/(2)-2x+2 and the ta...

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  18. The area of the closed figure bounded by y=1 //cos^(2)x,x=0,y=0and x=p...

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  19. The area (in square units) of the closed figure bounded by x=-1,x=2and...

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  20. The area bounded by y = 2-|2-x| and y=3/|x| is:

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