How do you find the circumference of the ellipse #x^2+4y^2=1#?

Question

Paisley Johnson · Accepted Answer

Using numerical techniques, we can get a approximation for this as:

# C = 4.8442 #

Although this seems like quite a simple question, the answer is actually ridiculous complicated. We need to first put the ellipse equation in standard form: #x^2+4y^2=1# # :. (x/1)^2+(y/(1/2))^2=1# Comparing with the standard equation; # (x/a)^2+(y/b)^2=1# We can identify this as an ellipse with semi-major axis #b=1/2# and semi-minor axis #a=1#, and the eccentricity of the ellipse is given by: # e=sqrt(1-(b/a)^2)) # # \ = sqrt(1-((1/2)/1)^2) # # \ = sqrt(3/4) # # \ = 1/2sqrt(3) # Then the exact circumference is given by: # C=4aE(e) # where #E(e)# is a complete elliptic integral of the second kind, # E(e) = int_0^(pi/2) \ sqrt(1-e^2sin^2 theta) \ d theta # Using numerical techniques, we can get a approximation for this as: # C = 4.8442 #

Christopher Allers · Answer

undefined To find the circumference of the ellipse $x^2 + 4y^2 = 1$, you can use the following formula:

$$
C = 4aE(e)
$$

where $a$ is the semi-major axis, $b$ is the semi-minor axis, and $e$ is the eccentricity of the ellipse. In this ellipse, since it's of the form $x^2/a^2 + y^2/b^2 = 1$, the semi-major axis $a$ is $1$ (since $a$ corresponds to the term with $x^2$) and the semi-minor axis $b$ is $\frac{1}{2}$ (since $b$ corresponds to the term with $y^2$).

To find $e$, the eccentricity, use the formula:

$$
e = \sqrt{1 - \frac{b^2}{a^2}}
$$

Substitute the values of $a$ and $b$ into the formula to find $e$. Then, plug in $a$ and $e$ into the formula $C = 4aE(e)$ to get the circumference of the ellipse.