Conic Sections Reference

Hyperbolas

A hyperbola has two branches and two asymptotes. The positive squared term tells us which direction the branches open, and the denominators build the rectangle used to draw the asymptotes.

Solved Examples

Jump directly to a worked example by problem type.

Summary Table

Feature How to Read It Why It Matters
Positive $x$ term The hyperbola opens left and right. The transverse axis is horizontal.
Positive $y$ term The hyperbola opens up and down. The transverse axis is vertical.
Center Read $(h,k)$ from the shifted terms. Everything is measured from this point.
Vertices Move $a$ units from the center along the opening direction. The vertices mark the closest points of the branches.
Asymptotes Use the rectangle with half-widths $a$ and $b$. The branches approach these lines but do not cross them in the basic graph.
Foci Move $c$ units from the center along the transverse axis. For hyperbolas, $c^2=a^2+b^2$.

Equation Forms

Horizontal Hyperbola
$$\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1$$
Vertical Hyperbola
$$\frac{(y-k)^2}{a^2}-\frac{(x-h)^2}{b^2}=1$$
Center
$$(h,k)$$
Focal Relationship
$$c^2=a^2+b^2$$
In a hyperbola, $a^2$ is always under the positive term in standard form. It marks the vertex distance from the center.

Asymptote Formulas

Horizontal Hyperbola
$$y-k=\pm\frac{b}{a}(x-h)$$
Vertical Hyperbola
$$y-k=\pm\frac{a}{b}(x-h)$$
Horizontal Vertices
$$(h-a,k),\quad(h+a,k)$$
Vertical Vertices
$$(h,k-a),\quad(h,k+a)$$
The asymptote slope comes from the guide rectangle: rise over run is $\frac ba$ for a horizontal hyperbola and $\frac ab$ for a vertical hyperbola.

Classic Examples

Read a Horizontal Hyperbola

Find the center, vertices, asymptotes, and foci of $\frac{(x-1)^2}{9}-\frac{(y+2)^2}{4}=1$.

Solution Steps
  1. Read the center and squared denominators.
  2. Take the positive square roots to find $a$ and $b$.
  3. State the center.
  4. Use the horizontal vertex pattern.
  5. Substitute the center and $a$ to find the vertices.
  6. State the vertices.
  7. Use the horizontal asymptote formula.
  8. Substitute the center, $a$, and $b$.
  9. State the asymptote equations.
  10. Use $c^2=a^2+b^2$.
  11. Evaluate $c$.
  12. Use the horizontal focus pattern.
  13. State the foci.
$$h=1,\quad k=-2,\quad a^2=9,\quad b^2=4$$ $$a=3,\quad b=2$$ $$\text{center: }(1,-2)$$ $$\text{vertices: }(h\pm a,k)$$ $$=(1\pm3,-2)$$ $$=(-2,-2),\ (4,-2)$$ $$\text{asymptotes: }y-k=\pm\frac ba(x-h)$$ $$y+2=\pm\frac23(x-1)$$ $$c^2=a^2+b^2$$ $$=9+4$$ $$c=\sqrt{13}$$ $$\text{foci: }(h\pm c,k)$$ $$=(1\pm\sqrt{13},-2)$$

Read a Vertical Hyperbola

Find the center, vertices, and asymptotes of $\frac{(y-3)^2}{16}-\frac{(x+1)^2}{9}=1$.

Solution Steps
  1. Read the center and squared denominators.
  2. Take the positive square roots to find $a$ and $b$.
  3. State the center.
  4. Use the vertical vertex pattern.
  5. Substitute the center and $a$ to find the vertices.
  6. State the vertices.
  7. Use the vertical asymptote formula.
  8. Substitute the center, $a$, and $b$.
$$h=-1,\quad k=3,\quad a^2=16,\quad b^2=9$$ $$a=4,\quad b=3$$ $$\text{center: }(-1,3)$$ $$\text{vertices: }(h,k\pm a)$$ $$=(-1,3\pm4)$$ $$=(-1,-1),\ (-1,7)$$ $$\text{asymptotes: }y-k=\pm\frac ab(x-h)$$ $$y-3=\pm\frac43(x+1)$$

Graphing Checklist

Step Action
1 Put the equation in standard form with $1$ on the right.
2 Find the center $(h,k)$.
3 Use the positive term to decide whether the hyperbola opens left-right or up-down.
4 Find $a$ and $b$ from the denominators.
5 Draw the guide rectangle and asymptote lines through the center.
6 Place the vertices and sketch branches approaching the asymptotes.