हिंदी

Sketch the curve {(x, y) : 100x2 + 25y2 = 2500} and find the area of the region enclosed by it, using integration.

Advertisements
Advertisements

प्रश्न

Sketch the curve {(x, y) : 100x2 + 25y2 = 2500} and find the area of the region enclosed by it, using integration.

योग
Advertisements

उत्तर

Divide the entire equation by 2500 to bring it into the standard form:

`(100x^2)/(2500) + (25y^2)/(2500) = 2500/2500`

`x^2/25 + y^2/100 = 1`

⇒ `x^2/5^2 + y^2/10^2 = 1`

The ellipse is centred at the origin (0, 0). To sketch it, plot the four intercepts:

(5, 0), (−5, 0) are the points of intersection of the curve and the x-axis.

(0, 10), (0, −10) are the points of intersection of the curve and the y-axis.

The area of the ellipse is 4 times the area in the first quadrant (x from 0 to 5).

From the equation:

`y^2/100 = 1 - x^2/25`

⇒ y = `10/5 sqrt(25 - x^2)`

= `2 sqrt(25 - x^2)`

Total Area = `4 int_0^5 ydx = 4 int_0^5 2sqrt(25 - x^2) dx`

= `8 int_0^5 sqrt(5^2 - x^2) dx`

Using the standard integral formula:

`int sqrt(a^2 - x^2) dx = x/2 sqrt(a^2 - x^2) + a^2/2 sin^-1 (x/a):`

Area = `8 [x/2 sqrt(25 - x^2) + 25/2 sin^-1 (x/5)]_0^5`

= `8 [(0 + 25/2 sin^-1 (1)) - (0 + 0)]`

= `8 [25/2 . pi/2]`

= `8 . (25pi)/4`

Area = 50π sq. units

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
2025-2026 (March) 65/5/1
Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×