# E-L-L-I-P-S-E

How to find the perimeter of an ellipse???

Note by Anirudha Nayak
4 years ago

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Let the ellipse be : $$\displaystyle \frac{x^2}{a^2}+\frac{y^2}{b^2} = 1$$

Parametric coordinates : $$\displaystyle (x,y) = (a\cos\theta,b\sin\theta)$$

$ds^2 = dx^2+dy^2$

$ds^2 = \left((\frac{dx}{d\theta})^2 + (\frac{dy}{d\theta})^2\right)d\theta^2$

$ds^2 = (a^2\sin^2\theta+ b^2\cos^2\theta)d\theta^2$

$ds = \sqrt{(a^2\sin^2\theta+ b^2\cos^2\theta)}d\theta$

Integrate it from $$\displaystyle 0$$ to $$\displaystyle 2\pi$$, and you will get the result.

- 4 years ago

Good luck integrating that monster.! (it hasn't been solved yet)

- 4 years ago

i am surprised This question came in VIT couldnt solve it maths was at a good level indeed when is urs or have u already given it

- 4 years ago

It was yesterday. What was the exact question?

- 4 years ago

i have posted it

- 4 years ago

how was d exam wht r u expecting

- 4 years ago

atleast 100 to aana chahiye.

- 4 years ago

me tooo simple paper though

- 4 years ago

ans for d question

- 4 years ago

Were there any options? I am asking because as Anish said that the value of the integral has no clear form. Maybe they intended to ask something different??

- 4 years ago

$$\pi(2^{1/2}),2\pi(2^{1/2}),\pi(2+2^{1/2})$$ and one other i dont remember them clearly

- 4 years ago

There is no exact formula but use the famous Indian mathematician Ramanujan came up with this better approximation:search Ramanujan circumference of ellipse formula

- 4 years ago