IMOClass 11 › Ellipse

Ellipse

Standard Equation and Elements

An ellipse is the set of points the sum of whose distances from two fixed points (the foci) is constant. A hyperbola is the set of points the difference of whose distances from two foci is constant. One word changes — sum versus difference — and a closed oval becomes a pair of opening branches.

Ellipse (foci on the $x$-axis, $a > b$):

$$\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1, \qquad a > b > 0$$

Here $2a$ is the major axis and $2b$ the minor axis. The foci sit at $(\pm c, 0)$ where $c = \sqrt{a^2 - b^2}$, and the vertices (endpoints of the major axis) are $(\pm a, 0)$.

Hyperbola (foci on the $x$-axis):

$$\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$$

The vertices are $(\pm a, 0)$, the foci are $(\pm c, 0)$ with $c = \sqrt{a^2 + b^2}$, and the two branches approach the slanted asymptotes $y = \pm \dfrac{b}{a}x$.

For both curves the shape is summarised by the eccentricity $e = \dfrac{c}{a}$, and both share the same latus-rectum formula:

$$e = \dfrac{c}{a}, \qquad \text{Length of latus rectum} = \dfrac{2b^2}{a}$$
ParameterEllipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$Hyperbola $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$
Defining conditionsum of focal distances $= 2a$difference of focal distances $= 2a$
Vertices$(\pm a, 0)$$(\pm a, 0)$
Foci$(\pm c, 0)$$(\pm c, 0)$
Relation$c = \sqrt{a^2 - b^2}$$c = \sqrt{a^2 + b^2}$
Eccentricity$0 < e < 1$$e > 1$
Latus rectum$\dfrac{2b^2}{a}$$\dfrac{2b^2}{a}$

The lone difference in the $c$-relations — a minus for the ellipse, a plus for the hyperbola — is exactly what pushes the eccentricity below $1$ in one case and above $1$ in the other.

Deeper Insight — eccentricity is the dial that turns one conic into another: Every conic in this chapter is a slice of a double cone, and a single number, the eccentricity $e$, records how steeply the slicing plane is tilted. A circle is the perfectly level cut with $e = 0$; tilt a little and you get an ellipse with $0 < e < 1$; tilt until the plane runs parallel to the cone's side and the closed oval breaks open into a parabola with $e = 1$; tilt further still and you cut both nappes of the cone, producing a hyperbola with $e > 1$. This is why ellipses, parabolas and hyperbolas are not three unrelated curves but three settings of the same dial — and why $e$ appears in the focus-directrix description of all of them ($PF = e \cdot PM$). The sign flip in $c = \sqrt{a^2 \mp b^2}$ is the algebraic shadow of this geometry: for an ellipse the foci must sit inside, so $c < a$ and $e < 1$; for a hyperbola the foci lie beyond the vertices, so $c > a$ and $e > 1$. Hold on to the cone picture and the entire chapter reads as one story rather than a list of formulae.

Ellipse with foci, vertices and axes labelled Ellipse: x²/a² + y²/b² = 1 (−a,0) (a,0) (−c,0) (c,0) major 2aminor 2b Hyperbola with two branches, foci, vertices and asymptotes Hyperbola: x²/a² − y²/b² = 1 asymptote (−a,0) (a,0) (−c,0) (c,0)
Example 1: Find the foci, vertices and eccentricity of the ellipse $\dfrac{x^2}{25} + \dfrac{y^2}{9} = 1$.
  1. Here $a^2 = 25$, $b^2 = 9$, so $a = 5$, $b = 3$ (major axis along the $x$-axis).
  2. $c = \sqrt{a^2 - b^2} = \sqrt{25 - 9} = \sqrt{16} = 4$.
  3. Vertices: $(\pm 5, 0)$; foci: $(\pm 4, 0)$.
  4. Eccentricity $e = \dfrac{c}{a} = \dfrac{4}{5}$.

Answer: Vertices $(\pm 5, 0)$, foci $(\pm 4, 0)$, $e = \dfrac{4}{5}$.

Example 2: Find the foci, vertices and eccentricity of the hyperbola $\dfrac{x^2}{16} - \dfrac{y^2}{9} = 1$.
  1. Here $a^2 = 16$, $b^2 = 9$, so $a = 4$, $b = 3$.
  2. For a hyperbola $c = \sqrt{a^2 + b^2} = \sqrt{16 + 9} = \sqrt{25} = 5$.
  3. Vertices: $(\pm 4, 0)$; foci: $(\pm 5, 0)$.
  4. Eccentricity $e = \dfrac{c}{a} = \dfrac{5}{4}$.

Answer: Vertices $(\pm 4, 0)$, foci $(\pm 5, 0)$, $e = \dfrac{5}{4}$.

Example 3: Find the length of the latus rectum of the ellipse $\dfrac{x^2}{36} + \dfrac{y^2}{20} = 1$.
  1. Here $a^2 = 36$, $b^2 = 20$, so $a = 6$.
  2. Length of latus rectum $= \dfrac{2b^2}{a} = \dfrac{2(20)}{6} = \dfrac{40}{6} = \dfrac{20}{3}$.

Answer: $\dfrac{20}{3}$.

Example 4: Find the equation of the ellipse with vertices $(\pm 6, 0)$ and foci $(\pm 4, 0)$.
  1. Vertices $(\pm a, 0)$ give $a = 6$, so $a^2 = 36$.
  2. Foci $(\pm c, 0)$ give $c = 4$, and $b^2 = a^2 - c^2 = 36 - 16 = 20$.
  3. The equation is $\dfrac{x^2}{36} + \dfrac{y^2}{20} = 1$.

Answer: $\dfrac{x^2}{36} + \dfrac{y^2}{20} = 1$.

Example 5: Find the equations of the asymptotes of the hyperbola $\dfrac{x^2}{9} - \dfrac{y^2}{4} = 1$.
  1. Here $a^2 = 9$, $b^2 = 4$, so $a = 3$, $b = 2$.
  2. The asymptotes of $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$ are $y = \pm \dfrac{b}{a}x$.
  3. Substitute: $y = \pm \dfrac{2}{3}x$.

Answer: $y = \dfrac{2}{3}x$ and $y = -\dfrac{2}{3}x$.

Example 6: Find the equation of the hyperbola with foci $(\pm 5, 0)$ and eccentricity $\dfrac{5}{3}$.
  1. Foci $(\pm c, 0)$ give $c = 5$; eccentricity $e = \dfrac{c}{a} = \dfrac{5}{3}$ gives $a = \dfrac{c}{e} = \dfrac{5}{5/3} = 3$, so $a^2 = 9$.
  2. For a hyperbola $b^2 = c^2 - a^2 = 25 - 9 = 16$.
  3. The equation is $\dfrac{x^2}{9} - \dfrac{y^2}{16} = 1$.

Answer: $\dfrac{x^2}{9} - \dfrac{y^2}{16} = 1$.

Quick recap
  • Ellipse: sum of focal distances is constant; $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ with $c=\sqrt{a^2-b^2}$ and $0
  • Hyperbola: difference of focal distances is constant; $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$ with $c=\sqrt{a^2+b^2}$ and $e>1$.
  • Both have vertices $(\pm a,0)$, foci $(\pm c,0)$, eccentricity $e=\dfrac{c}{a}$ and latus rectum $\dfrac{2b^2}{a}$.
  • Hyperbola asymptotes are $y=\pm\dfrac{b}{a}x$; the ellipse has none.
  • All conics are cone sections; the eccentricity $e$ (0, <1, =1, >1) names circle, ellipse, parabola, hyperbola.
✓ Quick check
The length of the minor axis of x²/49 + y²/16 = 1 is:
Minor axis length = 2b = 8.
Calculate the distance between the foci of x²/25 + y²/9 = 1.
a=5, e = √(1 - 9/25) = 4/5. Distance = 2ae = 2(5)(4/5) = 8.

Eccentricity and Latus Rectum

b² = a²(1 − e²) gives the eccentricity e (0 < e < 1). The latus rectum length is 2b²/a.

Example 1: Eccentricity of x²/25 + y²/9 = 1.
9 = 25(1 − e²) → e = 4/5.
Example 2: Latus rectum length there.
2·9/5 = 18/5.
Quick recap
  • b² = a²(1 − e²), 0 < e < 1.
  • Latus rectum = 2b²/a.
✓ Quick check
The eccentricity of x²/25 + y²/9 = 1 is:
e = c/a = 4/5.
Which point is a vertex of x²/64 + y²/49 = 1?
a = 8, so vertices are (±8,0).

Foci and Properties

The foci are at (±ae, 0) and the sum of focal distances of any point equals 2a.

Example 1: Foci of x²/25 + y²/9 = 1.
e = 4/5, ae = 4, foci (±4, 0).
Example 2: Sum of focal distances of any point on it.
2a = 10.
Quick recap
  • Foci at (±ae, 0).
  • Sum of focal distances = 2a.
✓ Quick check
Riya draws an ellipse with a = 10 and b = 6. Its eccentricity is:
c = √64 = 8, so e = 8/10 = 0.8.
In a whispering gallery in a Delhi museum shaped like a semi-ellipse, the length is 100 m and height is 30 m. How far are the foci from the center?
2a=100 => a=50. b=30. c = √(a² - b²) = √(2500 - 900) = √1600 = 40. Distance is 40 m.
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