मराठी

Find the area of the region bounded by the parabola y2 = 2px, x2 = 2py - Mathematics

Advertisements
Advertisements

प्रश्न

Find the area of the region bounded by the parabola y2 = 2px, x2 = 2py

बेरीज

उत्तर


We are given that: x2 = 2py   ......(i)

And y2 = 2px    ......(ii)

From equation (i)

We get y = `x^2/(2"p")`

Putting the value of y in equation (ii)

We have `(x^2/(2"p"))` = 2px

⇒ `x^4/(4"p"^2)` = 2px

⇒ x4 = 8p3x

⇒ x4 – 8p3x = 0

⇒ x(x3 – 8p3) = 0

∴ x = 0, 2p

Required area = Area of the region (OCBA – ODBA)

= `int_0^(2"p") sqrt(2"p"x)  "d"x - int_0^(2"p") x^2/(2"p")  "d"x`

= `sqrt(2"p") * 2/3 [x^(3/2)]_0^(2"p") - 1/(2"p") * 1/3 [x^3]_0^(2"p")`

= `(2sqrt(2))/3 sqrt("p") [(2"p")^(3/2) - 0] - 1/(6"p") [(2"p")^3 - 0]`

= `(2sqrt(2))/3 sqrt("p") * 2sqrt(2) "p"^(3/2) - 1/(6"p") * 8"p"^3`

= `8/3 * "p"^2 - 8/6 "p"^2`

= `8/6 "p"^2`

= `4/3 "p"^2` sq.units

Hence, the required area = `4/3 "p"^2` sq.units

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 8: Application Of Integrals - Exercise [पृष्ठ १७६]

APPEARS IN

एनसीईआरटी एक्झांप्लर Mathematics [English] Class 12
पाठ 8 Application Of Integrals
Exercise | Q 2 | पृष्ठ १७६

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

Using integration, find the area bounded by the curve x2 = 4y and the line x = 4y − 2.


Find the area bounded by the curve y = sin x between x = 0 and x = 2π.


Sketch the graph of y = \[\sqrt{x + 1}\]  in [0, 4] and determine the area of the region enclosed by the curve, the x-axis and the lines x = 0, x = 4.


Find the area under the curve y = \[\sqrt{6x + 4}\] above x-axis from x = 0 to x = 2. Draw a sketch of curve also.


Sketch the region {(x, y) : 9x2 + 4y2 = 36} and find the area of the region enclosed by it, using integration.


Find the area of the region bounded by the curve xy − 3x − 2y − 10 = 0, x-axis and the lines x = 3, x = 4.


Draw a rough sketch of the curve \[y = \frac{x}{\pi} + 2 \sin^2 x\] and find the area between the x-axis, the curve and the ordinates x = 0 and x = π.


Using integration, find the area of the region bounded by the triangle ABC whose vertices A, B, C are (−1, 1), (0, 5) and (3, 2) respectively.


Find the area of the region common to the circle x2 + y2 = 16 and the parabola y2 = 6x.


Draw a rough sketch of the region {(x, y) : y2 ≤ 3x, 3x2 + 3y2 ≤ 16} and find the area enclosed by the region using method of integration.


Find the area of the region bounded by \[y = \sqrt{x}, x = 2y + 3\]  in the first quadrant and x-axis.


Find the area enclosed by the curve \[y = - x^2\] and the straight line x + y + 2 = 0. 


Using integration, find the area of the following region: \[\left\{ \left( x, y \right) : \frac{x^2}{9} + \frac{y^2}{4} \leq 1 \leq \frac{x}{3} + \frac{y}{2} \right\}\]


If the area bounded by the parabola \[y^2 = 4ax\] and the line y = mx is \[\frac{a^2}{12}\] sq. units, then using integration, find the value of m. 

 


Find the area bounded by the parabola x = 8 + 2y − y2; the y-axis and the lines y = −1 and y = 3.


Find the area of the region bounded by the curve ay2 = x3, the y-axis and the lines y = a and y = 2a.


Find the area of the region bounded by the curves y2 = 9x, y = 3x


Sketch the region `{(x, 0) : y = sqrt(4 - x^2)}` and x-axis. Find the area of the region using integration.


Compute the area bounded by the lines x + 2y = 2, y – x = 1 and 2x + y = 7.


Area of the region in the first quadrant enclosed by the x-axis, the line y = x and the circle x2 + y2 = 32 is ______.


Using integration, find the area of the region bounded between the line x = 4 and the parabola y2 = 16x.


Find the area bounded by the curve y = |x – 1| and y = 1, using integration.


The area enclosed by y2 = 8x and y = `sqrt(2x)` that lies outside the triangle formed by y = `sqrt(2x)`, x = 1, y = `2sqrt(2)`, is equal to ______.


Let g(x) = cosx2, f(x) = `sqrt(x)`, and α, β (α < β) be the roots of the quadratic equation 18x2 – 9πx + π2 = 0. Then the area (in sq. units) bounded by the curve y = (gof)(x) and the lines x = α, x = β and y = 0, is ______.


The area of the region bounded by the parabola (y – 2)2 = (x – 1), the tangent to it at the point whose ordinate is 3 and the x-axis is ______.


Find the area of the smaller region bounded by the curves `x^2/25 + y^2/16` = 1 and `x/5 + y/4` = 1, using integration.


Sketch the region bounded by the lines 2x + y = 8, y = 2, y = 4 and the Y-axis. Hence, obtain its area using integration.


Using integration, find the area of the region bounded by y = mx (m > 0), x = 1, x = 2 and the X-axis.


Make a rough sketch of the region {(x, y) : 0 ≤ y ≤ x2 + 1, 0 ≤ y ≤ x + 1, 0 ≤ x ≤ 2} and find the area of the region, using the method of integration.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×