Ellipses
Ellipses
The set of all points whose distances from two fixed points (foci) have a constant sum $2a$.
Horizontal major axis: $$\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1, \quad a > b$$
Vertical major axis: $$\frac{(x-h)^2}{b^2} + \frac{(y-k)^2}{a^2} = 1, \quad a > b$$
$$c^2 = a^2 - b^2$$
Vertices are $a$ units from center along major axis. Foci are $c$ units from center.
Eccentricity: $e = c/a$, where $0 < e < 1$.
Find the foci and vertices: $\frac{x^2}{25} + \frac{y^2}{9} = 1$.
$a^2 = 25$, $b^2 = 9$, $c^2 = 16$, $c = 4$. Horizontal major axis.
Vertices: $(\pm 5, 0)$. Foci: $(\pm 4, 0)$.
Find the eccentricity of $\frac{x^2}{100}+\frac{y^2}{64}=1$.
$c = \sqrt{100-64} = 6$. $e = 6/10 = 0.6$.
Write the equation of an ellipse centered at origin with vertices $(0,\pm 6)$ and foci $(0,\pm 4)$.
Vertical major axis: $a = 6$, $c = 4$, $b^2 = 36 - 16 = 20$.
$$\frac{x^2}{20} + \frac{y^2}{36} = 1$$
Practice Problems
Show Answer Key
1. Vertices $(\pm 4,0)$, $c=\sqrt{12}=2\sqrt{3}$; foci $(\pm 2\sqrt{3},0)$
2. $c = \sqrt{169-25} = 12$
3. $\frac{x^2}{64}+\frac{y^2}{9}=1$
4. $c = \sqrt{24} = 2\sqrt{6}$; $e = 2\sqrt{6}/7 \approx 0.70$
5. Vertical ($a^2 = 16$ is under $y^2$)
6. Major = $2a = 12$, Minor = $2b = 10$
7. $b^2 = 25-9 = 16$; $\frac{x^2}{25}+\frac{y^2}{16}=1$
8. Center $(1,-2)$, $a = 3$, $b = 2$
9. $c = 8$, $b = \sqrt{100-64} = 6$
10. A circle
11. $(0, \pm 3)$
12. $c = 0.017 \times 150 = 2.55$ million km