# Terminology Of Ellipses

Go back to  'Ellipse'

We now discuss some terminology related to an ellipse which we’ve already seen for the case of parabolas.

FOCAL DISTANCES OF Q (x, y): Let  \begin{align}\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1\end{align} be the ellipse and $$F, F'$$ be the two foci (refer to fig - 6). The focal distances of $$Q$$ are simply the two distances $$QF$$ and $$QF.$$

$QF = eQN = e\left( {x + \frac{a}{e}} \right) = a + ex$

$QF' = eQN' = e\left( {\frac{a}{e} - x} \right) = a - ex$

Thus, the two focal distances of any point $$(x, y)$$ are $$(a + ex)$$ and $$(a – ex).$$ The sum is $$2a,$$ as expected.

LATUS RECTUM: This is the chord passing through any of the two foci and perpendicular to the major axis. To evaluate the length of the latus-rectum, we must evaluate the y-coordinates of the two extremities of the latus rectum. In the equation of the ellipse

$\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$

we substitute $$x = \pm ae$$ ( the x-coordinates of the two foci). Thus

\begin{align}&{y^2} = {b^2}\left( {1 - \frac{{{x^2}}}{{{a^2}}}} \right)\\&{\rm{ \quad}} = {a^2}{(1 - {e^2})^2}\end{align}

\begin{align}&y = \pm a(1 - {e^2})\\&{\rm{ \;\;}} = \pm \frac{{{b^2}}}{a}\end{align}

Thus, the length $$l$$ of the latus-rectum is

$l = \frac{{2{b^2}}}{a}$

Example - 1

Find the lengths of the major and minor axes and the foci for these ellipses:

(a)   $$16{x^2} + 25{y^2} = 400$$

(b)   $${x^2} + 4{y^2} - 2x = 0$$

(c)   $$25{x^2} + 16{y^2} = 400$$

Solution: (a)  The given equation can be written in the form of the standard equation of an ellipse:

\begin{align}&\quad\frac{{16{x^2}}}{{400}} + \frac{{25{y^2}}}{{400}} = 1\\&\Rightarrow\quad \; \frac{{{x^2}}}{{25}} + \frac{{{y^2}}}{{16}} = 1\\&\Rightarrow\quad\; \frac{{{x^2}}}{{{5^2}}} + \frac{{{y^2}}}{{{4^2}}} = 1\end{align}

Thus, comparing with the standard form, we have

$$a = 5, \; b = 4$$

The major axis is of length $$2a = 10$$ and the minor axis is of length $$2b = 8.$$

The eccentricity of the ellipse is

$e = \sqrt {1 - \frac{{{b^2}}}{{{a^2}}}} = \frac{3}{5}$

Thus, the two foci are at $$( \pm ae,\,0),$$ i.e., at $$( \pm 3,\,\,0).$$

(b)  We first rearrange the equation of the ellipse to standard form :

\begin{align}&\qquad\qquad\quad{x^2} + 4{y^2} - 2x = 1\\&\Rightarrow \qquad\quad {(x - 1)^2} + 4{y^2} = 1\\&\Rightarrow \quad\frac{{{{(x - 1)}^2}}}{{{1^2}}} + \frac{{{y^2}}}{{{{(1/2)}^2}}} = 1\end{align}

Instead, of $$x,$$ we have $$x –1.$$ This means that the center of the ellipse is at $$(1, 0)$$ instead of $$(0, 0).$$ As in the unit on parabola, we can use a translation of the axes (refer to Parabolas):

$X \to x - 1,Y \to y.$

so that the equation of the ellipse in the $$X-Y$$ system is

$\frac{{{X^2}}}{{{1^2}}} + \frac{{{Y^2}}}{{{{\left( {1/2} \right)}^2}}} = 1$

Since $$a = 1$$ and \begin{align}b = \frac{1}{2},\end{align} the major and minor axes are 2 and 1 respectively, while the eccentricity is \begin{align}e = \sqrt {1 - \frac{1}{4}} = \frac{{\sqrt 3 }}{2}.\end{align} Thus, the two foci are at (in the X-Y system):

$\left( { \pm \frac{{\sqrt 3 }}{2},\,0} \right)$

In the original x-y system, we use the reverse transformation $$x \to X + 1,\,\,y \to Y,$$ so that the foci in the original system are at

$\left( {1 \pm \frac{{\sqrt 3 }}{2},\,0} \right)$

(c)  $$25{x^2} + 16{y^2} = 400$$

$\Rightarrow \qquad \frac{{{x^2}}}{{16}} + \frac{{{y^2}}}{{25}} = 1$

$\Rightarrow \quad \frac{{{x^2}}}{{{{(4)}^2}}} + \frac{{{y^2}}}{{{{(5)}^2}}} = 1$

Here, we see that $$b = 5$$ is greater than $$a = 4,$$ which means that the major axis will lie not along the $$x\text{-axis}$$ but along the $$y\text{-axis,}$$ and obviously, since the foci lie on the major axis, the foci will also lie on the $$y\text{-axis.}$$ The major and minor axes are of lengths $$2b = 10$$ and $$2a = 8$$ respectively. The eccentricity is \begin{align}e = \sqrt {1 - \frac{{{a^2}}}{{{b^2}}}} \end{align} (a and b get interchanged )\begin{align} = \frac{3}{5}.\end{align} Thus, the two foci lie at $$(0,\, \pm be),$$ i.e. at $$(0,\,\, \pm 3).$$ The length of the latus rectum in this case will be given by \begin{align}\frac{{2{a^2}}}{b}\end{align} and will be equal to \begin{align}\frac{{32}}{5}\end{align}

We pause briefly and summarize whatever we’ve covered uptill now.

 SUMMARY BASICS ON ELLIPSE Definition – 1 An ellipse is the locus of a moving point such that the ratio of its distance from a fixed point to its distance from a fixed line is a  constant less than unity. This constant is termed the eccentricity of the ellipse. The fixed point is the focus while the fixed line is the directrix. The symmetrical nature of the ellipse ensures that there will be two foci and two directrices. Definition – 2 An ellipse is the locus of a moving point such that the sum of its distances from two fixed points is constant. The two fixed points are the two foci of the ellipse. To plot the ellipse, we can use the peg-and-thread method described earlier. STANDARD EQUATION $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$ $\textbf{If }\; \bf{a > b}$ $\textbf{If }\; \bf{a < b}$ Vertices Foci Major axis Minor axis Directrices Eccentricity e Latus-rectum Focal distances of (x,y) $$(a, 0)$$ and $$(-a, 0)$$ $$(ae, 0)$$ and $$(-ae, 0)$$ $$2a$$ (along x-axis) $$2b$$ (along y-axis) \begin{align}x = \frac{a}{e}{\text{ and }} x = - \frac{a}{e}\end{align} \begin{align}\sqrt {1 - \frac{{{b^2}}}{{{a^2}}}} \end{align} \begin{align}\frac{{2{b^2}}}{a}\end{align} $$a \pm ex$$ $$(0, b)$$ and $$(0, -b)$$ $$(0, be)$$ and $$(0, -be)$$ $$2b$$ (along y-axis) $$2a$$ (along x-axis) \begin{align}y = \frac{b}{e}{\text{ and }}y = - \frac{b}{e} \end{align} \begin{align}\sqrt {1 - \frac{{{a^2}}}{{{b^2}}}} \end{align} \begin{align}\frac{{2{a^2}}}{b}\end{align} $$b \pm ey$$

And lastly, if the equation of the ellipse is

$\frac{{{{(x - \alpha )}^2}}}{{{a^2}}} + \frac{{{{(y - \beta )}^2}}}{{{b^2}}} = 1$

instead of the usual standard form, we can use the transformation $$X \to x - \alpha {\text{ and }}Y \to y - \beta$$ and (basically a translation of the axes so that the origin of the new system coincides with $$(\alpha ,\,\,\beta ).$$ The equation then becomes

$\frac{{{X^2}}}{{{a^2}}} + \frac{{{Y^2}}}{{{b^2}}} = 1$

We can now work on this form, use all the standard formulae that we’d like to and obtain whatever it is that we wish to obtain. The final result (in the x-y system) is obtained using the reverse transformation $$x \to X + \alpha {\text{ and }}y \to Y + \beta.$$