Question

Find the area of the ellipse `x^2/a^2 + y^2/b^2 = 1`.

Hence write area of `x^2/25 + y^2/16 = 1`.

Answer 1

Equation of ellipse `x^2/a^2 + y^2/b^2 = 1`

The area of the ellipse is 4 times the area of region OPQO, as shown in the figure. For the region, the limits of integration are x = 0 and x = a.

From the equation ellipse,

`x^2/a^2 + y^2/b^2 = 1`
∴ `y^2/b^2 = 1 - x^2/a^2`

∴ `y^2 = b^2.((a^2 - x^2)/a^2)`

∴ `y =  b/a. sqrt(a^2 - x^2)`   ...(∵ in 1st quadrant y > 0)

We know,

A = `4int_(x = 0)^a y. dx`

= `4 int_0^a b/a. sqrt(a^2 - x^2). dx`

= `(4b)/a [x/2 sqrt(a^2 - x^2) + a^2/2 sin^-1  x/a]_0^a` 

= `(4b)/a [{a/a sqrt(a^2 - a^2) + a^2/2 sin^(-1)  a/a} - {0/a sqrt(a^2 - 0^2) + a^2/2 sin^(-1)  0/a}]`

= `(4b)/a [(a^2/2. π/2) - 0]`

A = πab sq. units.    ...(1)

Now,

`x^2/25 + y^2/16 = 1`

∴ `x^2/5^2 + y^2/4^2 = 1`

Comparing `x^2/5^2 + y^2/4^2 = 1` with `x^2/a^2 + y^2/b^2 = 1`, we get, a = 5 b = 4

∴ Area of `x^2/25 + y^2/16 = 1` is 20π sq. units.   ...[From (1)]