The three conic sections

Planets circle the Sun along ellipses. A satellite dish collects signals at a single point because its cross-section is a parabola. A power-station cooling tower has the shape of a hyperbola turned around. These three curves, together with the circle, are the conic sections, and each has a short definition in terms of distances to special points.

The ellipse

With $F_1(-c, 0)$, $F_2(c, 0)$ and $b^2=a^2-c^2$ the equation is $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\qquad(a>b>0).$$ The vertices are $(\pm a, 0)$ and $(0, \pm b)$; $2a$ is the length of the major axis, $2b$ of the minor axis.

The hyperbola

With foci $(\pm c, 0)$ and $b^2=c^2-a^2$: $$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1.$$ The vertices are $(\pm a, 0)$ and the curve approaches the asymptotes $y=\pm\frac{b}{a}x$.

The parabola

With $F\left(\frac p2, 0\right)$ and $\Delta: x=-\frac p2$ ($p>0$) the equation is $$y^2=2px.$$

Using the definitions

Key terms

Summary

• Ellipse: $MF_1+MF_2=2a$, $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, $c^2=a^2-b^2$.
• Hyperbola: $|MF_1-MF_2|=2a$, $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$, $c^2=a^2+b^2$, asymptotes $y=\pm\frac bax$.
• Parabola: distance to $F$ equals distance to $\Delta$; $y^2=2px$, focus $\left(\frac p2, 0\right)$, directrix $x=-\frac p2$.

Practice

Show solutions

1. $a=7$, $b=5$, $c^2=49-25=24$, so $c=2\sqrt6$. Foci $(\pm2\sqrt6, 0)$; vertices $(\pm7, 0)$ and $(0, \pm5)$.
2. $c=2$, $a=3$, $b^2=9-4=5$: $\frac{x^2}{9}+\frac{y^2}{5}=1$.
3. $\frac{x^2}{9}+\frac{y^2}{4}=1$: $a=3$, $b=2$, $c=\sqrt5$, foci $(\pm\sqrt5, 0)$.
4. $a=5$, $b=12$, $c=\sqrt{25+144}=13$. Foci $(\pm13, 0)$; asymptotes $y=\pm\frac{12}{5}x$.
5. $a=2$, $c^2=13$, $b^2=13-4=9$: $\frac{x^2}{4}-\frac{y^2}{9}=1$.
6. $2p=8$, so $p=4$. Focus $(2, 0)$; directrix $x=-2$.
7. $\frac p2=\frac32$, so $p=3$: $y^2=6x$.
8. $\frac{25}{9}-\frac{(16/3)^2}{16}=\frac{25}{9}-\frac{256}{144}=\frac{25}{9}-\frac{16}{9}=1$ ✓. Foci $(\pm5, 0)$: $PF_2=\frac{16}{3}$ and $PF_1=\sqrt{100+\frac{256}{9}}=\frac{34}{3}$. Difference $\frac{18}{3}=6=2a$ ✓.
9. For this ellipse $MF_2=5-\frac45x$. Setting $5-0.8x=3$ gives $x=2.5$, then $y^2=9\left(1-\frac{6.25}{25}\right)=6.75$, $y=\pm\frac{3\sqrt3}{2}$. Points $\left(2.5, \pm\frac{3\sqrt3}{2}\right)$. Check: $\sqrt{1.5^2+6.75}=\sqrt9=3$ ✓.
10. Put the vertex at the origin with the axis along $Ox$: the rim point is $(0.5, 1)$. $1^2=2p\cdot0.5$ gives $p=1$, so the focus is at $\frac p2=0.5$ m from the vertex.