The hyperbola: a locus, its equation and asymptotes

A hyperbola is the locus of all the points in the plane whose distances from two given points, the foci, differ by a constant. The constant is called $\displaystyle 2a$, and it is less than the distance $\displaystyle 2c$ between the foci.

With the foci $\displaystyle F_1(-c,0)$ and $\displaystyle F_2(c,0)$ the condition is $\displaystyle {|PF_1-PF_2|}={2a}$. The same steps as for the ellipse – isolate a root and square twice – lead to $\displaystyle {(c^{2}-a^{2})x^{2}-a^{2}y^{2}}={a^{2}(c^{2}-a^{2})}$. This time $\displaystyle c>a$, and we write $\displaystyle {b^{2}}={c^{2}-a^{2}}$. This gives the canonical equation of the hyperbola:

• Symmetry about both axes and the origin.
• The vertices $\displaystyle (\pm a,0)$. The $\displaystyle y$-axis is not met: $\displaystyle {x}={0}$ gives $\displaystyle {y^{2}}={-b^{2}}$.
• Two branches: $\displaystyle {\frac{x^{2}}{a^{2}}}={1+\frac{y^{2}}{b^{2}}\ge1}$, so $\displaystyle x\ge a$ (the right branch, near $\displaystyle F_2$) or $\displaystyle x\le-a$ (the left branch).

Isolate $\displaystyle y$: $\displaystyle {y}={\pm\frac{b}{a}\sqrt{x^{2}-a^{2}}}$. When $\displaystyle x$ is very large, $\displaystyle \sqrt{x^{2}-a^{2}}$ is almost $\displaystyle |x|$, and the hyperbola comes close to the lines

– the asymptotes of the hyperbola. They pass through the origin, and the hyperbola comes closer and closer to them without meeting them.

An example: $\displaystyle {\frac{x^{2}}{16}-\frac{y^{2}}{9}}={1}$: $\displaystyle {a}={4}$, $\displaystyle {b}={3}$, $\displaystyle {c}={\sqrt{16+9}}={5}$.

• The vertices are $\displaystyle (\pm4,0)$, the foci $\displaystyle (\pm5,0)$, and the asymptotes $\displaystyle {y}={\pm\frac{3}{4}x}$.
• A check at the vertex $\displaystyle (4,0)$: its distances from the foci are $\displaystyle 9$ and $\displaystyle 1$, and the difference is $\displaystyle {8}={2a}$.

When the foci are on the $\displaystyle y$-axis, the signs swap: $\displaystyle {-\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}}={1}$.

• The vertices are $\displaystyle (0,\pm b)$, the branches go up and down, and the foci are $\displaystyle (0,\pm c)$ with $\displaystyle {c}={\sqrt{a^{2}+b^{2}}}$.
• The asymptotes are the same: $\displaystyle {y}={\pm\frac{b}{a}x}$.

For example, $\displaystyle {-\frac{x^{2}}{9}+\frac{y^{2}}{16}}={1}$: the vertices are $\displaystyle (0,\pm4)$ and the foci $\displaystyle (0,\pm5)$.

The hyperbola: $\displaystyle {\frac{x^{2}}{9}-\frac{y^{2}}{16}}={1}$.

What are the vertices and the foci?

Write the two asymptotes on the board.

A hyperbola’s foci are $\displaystyle (\pm5,0)$, and every point’s distances from the foci differ by $\displaystyle 6$.
Write the equation of the hyperbola.

The hyperbola: $\displaystyle {9x^{2}-4y^{2}}={36}$.

Write the equation in canonical form. What are the vertices and the asymptotes?

A hyperbola’s asymptotes are $\displaystyle {y}={\pm2x}$, and one of its vertices is $\displaystyle (1,0)$.
Write the equation of the hyperbola.

The point $\displaystyle P(8,y_P)$, with $\displaystyle y_P>0$, is on the hyperbola $\displaystyle {\frac{x^{2}}{16}-\frac{y^{2}}{9}}={1}$.
$\displaystyle F_1(-5,0)$ and $\displaystyle F_2(5,0)$ are the foci.