Foci, vertices, asymptotes and eccentricity

From the canonical equation $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$ one immediately obtains all the elements needed to draw the curve.

Property — Elements of the hyperbola

For the hyperbola $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$:

• Foci: $F_{1,2}(\mp c; 0)$ with $c = \sqrt{a^2 + b^2}$.
• Vertices: $V_{1,2}(\mp a; 0)$. On the $y$-axis there are no vertices: setting $x = 0$ the equation becomes $-\tfrac{y^2}{b^2} = 1$, which is impossible.
• Asymptotes: $y = \pm \tfrac{b}{a}\,x$, two lines through the origin which the hyperbola approaches indefinitely.
• Eccentricity: $e = c/a$, with $e > 1$.

Unlike the ellipse, the hyperbola meets only one of the two axes (the one on which the foci lie) and is made up of two distinct branches that open outwards along the asymptotes.

Reading: the hyperbola $\tfrac{x^2}{4}-\tfrac{y^2}{3}=1$ has vertices $V_{1,2}(\mp 2;0)$, foci $F_{1,2}(\mp\sqrt{7};0)\approx(\mp 2{,}83;0)$ and asymptotes $y=\pm\tfrac{\sqrt{3}}{2}x$ (the two dashed lines that the branches approach).

In the following simulation you can vary the semi-axes $a$ and $b$ and watch how the asymptotes, foci and eccentricity change.

Drag the sliders $a$ and $b$: the asymptotes $y=\pm\tfrac{b}{a}x$, the foci $F_{1,2}(\mp c;0)$ and the eccentricity $e=c/a$ update.

Links

Topics: Homographic hyperbola
Concepts: Asymptote · Eccentricity · Foci · Hyperbola · Vertices
Functions: Hyperbola
Skills: Analytic geometry · Sketching the graph