Elements of Ellipse

Elements of the ellipse are shown in the figure below.
 

Elements of ellipse

 

  1. Center (h, k). At the origin, (h, k) is (0, 0).
  2. Semi-major axis = a and semi-minor axis = b.
  3. Location of foci c, with respect to the center of ellipse. $c = \sqrt{a^2 - b^2}$.
  4. Length latus rectum, LR
    semi-latus rectum of ellipseConsider the right triangle F1QF2:
  5. 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
    $(2c)^2 + (\frac{1}{2}LR)^2 = z^2$

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

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

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

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

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

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

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

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

    $LR = \dfrac{2b^2}{a}$

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

  6. 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,
    $e = \dfrac{a}{d} = \dfrac{a}{a^2 / c}$

    $e = \dfrac{c}{a} \lt 1.0$

     

  7. Location of directrix d, with respect to the center of ellipse.

    From the derivation of eccentricity,

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

 

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