# Elements of Ellipse

Elements of the ellipse are shown in the figure below.

- Center (
*h*,*k*). At the origin, (*h*,*k*) is (0, 0). - Semi-major axis =
*a*and semi-minor axis =*b*. - Location of foci
*c*, with respect to the center of ellipse. $c = \sqrt{a^2 - b^2}$. - Length latus rectum,
*LR*

Consider the right triangle F_{1}QF_{2}: - Eccentricity,
*e*

$e = \dfrac{\text{distance from focus to ellipse}}{\text{distance from ellipse to directrix}}$

From the figure of the ellipse above,

$e = \dfrac{d_3}{d_4} = \dfrac{a}{d} = \dfrac{a - c}{d - a}$From

$\dfrac{a}{d} = \dfrac{a - c}{d - a}$$ad - a^2 = ad - cd$

$d = a^2 / c$

Thus,

$e = \dfrac{a}{d} = \dfrac{a}{a^2 / c}$$e = \dfrac{c}{a} \lt 1.0$ - Location of directrix
*d*, with respect to the center of ellipse.From the derivation of eccentricity,

$d = \dfrac{a}{e} \, \text{ or } d = \dfrac{a^2}{c}$

Based on the definition of ellipse

$z + \frac{1}{2}LR = 2a$

$z = 2a - \frac{1}{2}LR$

$z = \dfrac{4a - LR}{2}$

By Pythagorean Theorem

$(2c)^2 + (\frac{1}{2}LR)^2 = z^2$

$4c^2 + \frac{1}{4}(LR)^2 = \left( \dfrac{4a - LR}{2} \right)^2$

$4c^2 + \frac{1}{4}(LR)^2 = \dfrac{(4a - LR)^2}{4}$

$16c^2 + (LR)^2 = (4a - LR)^2$

$16c^2 + (LR)^2 = 16a^2 - 8a(LR) + (LR)^2$

$16c^2 = 16a^2 - 8a(LR)$

$8a(LR) = 16a^2 - 16c^2$

$LR = \dfrac{16a^2 - 16c^2}{8a}$

$LR = \dfrac{16(a^2 - c^2)}{8a}$

You can also find the same formula for the length of latus rectum of ellipse by using the definition of eccentricity.