Advertisements
Advertisements
प्रश्न
Find the area enclosed between the parabola 4y = 3x2 and the straight line 3x - 2y + 12 = 0.
Advertisements
उत्तर १
The given quations are
`y = (3x^2)/4` ....1
and, 3x – 2y + 12 = 0 ...(2)
Solving equation no. (1) & (2)
y = `(3x + 12)/2`
उत्तर २
The equations of the curves are
4y = 3x2 .....(1)
3x – 2y + 12 = 0 .....(2)
The curve (1) represents a parabola having vertex at the origin, axis along the positive direction of y-axis and opens upwards.
The curve (2) represents a straight line. This straight line meets the x-axis at (−4, 0) and y-axis at (0, 6).
Soving (1) and (2), we get
`3x - 2((3x^2)/4) + 12 = 0`
⇒ 6x - 3x2 + 24 = 0
⇒ x2 - 2x−8=0
⇒ (x+2)(x−4) = 0
⇒ x = -2, 4
When x = − 2, y = 3
When x = 4, y = 12
So, the point of intersection of the given curves is (− 2, 3) and (4, 12).
The area bounded by the given curves is shown below.
Required area = Area of the shaded region
= `int_(-2)^4 (y"Line" - y"Parabola") dx`
= `int_(-2)^4 ((3x + 12)/2 - 3/4 x^2) dx`
`= (3/4 x^2 + 6x - x^3/4)^4`
`= (3/4 xx 16 + 6 xx 4 - 64/4) - (3/4 xx 4 +6 xx (-2) + 3/4)`
= 20 - (- 7)
= 27 square units
संबंधित प्रश्न
Find the area of the region bounded by y2 = 9x, x = 2, x = 4 and the x-axis in the first quadrant.
Find the area of the smaller part of the circle x2 + y2 = a2 cut off by the line `x = a/sqrt2`
Find the area of the region bounded by the curve y2 = 4x and the line x = 3
Find the area under the given curve and given line:
y = x2, x = 1, x = 2 and x-axis
Find the area enclosed by the parabola 4y = 3x2 and the line 2y = 3x + 12
Find the area of the smaller region bounded by the ellipse `x^2/a^2 + y^2/b^2 = 1` and the line `x/a + y/b = 1`
Using the method of integration find the area of the triangle ABC, coordinates of whose vertices are A(2, 0), B (4, 5) and C (6, 3).
Find the area of the region bounded by the following curves, the X-axis and the given lines: y = `sqrt(16 - x^2)`, x = 0, x = 4
Find the area of the region bounded by the following curves, the X-axis and the given lines: 2y = 5x + 7, x = 2, x = 8
Choose the correct alternative :
Area of the region bounded by the curve x2 = y, the X-axis and the lines x = 1 and x = 3 is _______.
If the curve, under consideration, is below the X-axis, then the area bounded by curve, X-axis and lines x = a, x = b is positive.
Area of the region bounded by the curve x = y2, the positive Y axis and the lines y = 1 and y = 3 is ______
Choose the correct alternative:
Area of the region bounded by y2 = 16x, x = 1 and x = 4 and the X axis, lying in the first quadrant is ______
State whether the following statement is True or False:
The equation of the area of the circle is `x^2/"a"^2 + y^2/"b"^2` = 1
The area of the region bounded by the curve y2 = x and the Y axis in the first quadrant and lines y = 3 and y = 9 is ______
The area of the region bounded by y2 = 25x, x = 1 and x = 2 the X axis is ______
Find the area of the region bounded by the parabola y2 = 25x and the line x = 5
Find area of the region bounded by 2x + 4y = 10, y = 2 and y = 4 and the Y-axis lying in the first quadrant
Find area of the region bounded by the curve y = – 4x, the X-axis and the lines x = – 1 and x = 2
Find the area of the region bounded by the curve y = `sqrt(36 - x^2)`, the X-axis lying in the first quadrant and the lines x = 0 and x = 6
`int_0^log5 (e^xsqrt(e^x - 1))/(e^x + 3)` dx = ______
Area enclosed between the curve y2(4 - x) = x3 and line x = 4 above X-axis is ______.
The area of the circle `x^2 + y^2 = 16`, exterior to the parabola `y = 6x`
Find the area between the two curves (parabolas)
y2 = 7x and x2 = 7y.
The area of the region bounded by the curve y = sin x and the x-axis in [–π, π] is ______.
Area bounded by y = sec2x, x = `π/6`, x = `π/3` and x-axis is ______.
The figure shows as triangle AOB and the parabola y = x2. The ratio of the area of the triangle AOB to the area of the region AOB of the parabola y = x2 is equal to ______.
