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Math101learn.math101.caHyperbola
A hyperbola is the set of points for which the absolute difference of distances to two foci is constant.
Hyperbolas model navigation time differences, telescope mirrors, cooling towers, and reciprocal-type behaviour. Their asymptotes connect exact loci with limiting geometry.
Intuition and core definition
A hyperbola is the set of points for which the absolute difference of distances to two foci is constant. Standard forms have one positive and one negative squared term; the positive term identifies the direction the two branches open.
Notation, language, and conditions
Horizontal form is $(x-h)^2/a^2-(y-k)^2/b^2=1$; vertical form reverses the terms. The centre is $(h,k)$, vertices lie $a$ units along the transverse axis, foci lie $c$ units with $c^2=a^2+b^2$, and asymptotes guide branch behaviour.
Why this idea matters
A hyperbola consists of two branches governed by a constant difference of focal distances and approached by asymptote lines.
A dependable method
- Rewrite the equation in standard form equal to $1$.
- Read the centre and identify the positive squared term.
- Take square roots to obtain $a$ and $b$.
- Find vertices and compute foci using $c^2=a^2+b^2$.
- Draw the guiding rectangle and asymptotes, then sketch branches through vertices.
Worked example
Representations and interpretation
A rectangle of half-width $a$ and half-height $b$ supplies asymptote diagonals. The graph approaches these lines without becoming them, while the focal-distance difference remains $2a$.
Reasoning about variations
Unlike an ellipse, a hyperbola uses $c^2=a^2+b^2$, so its foci lie beyond the vertices. Swapping which squared term is positive rotates the opening direction by $90^\circ$.
Common mistakes
How to check your work
- Substitute vertices into the equation.
- Confirm $c>a$ and foci lie outside vertices.
- Check asymptote slopes against the guiding rectangle.
Practice
- For $x^2/16-y^2/9=1$, find $c$.
- Which way does $y^2/4-x^2/25=1$ open?
- Find vertices of $(x-2)^2/9-(y+1)^2/4=1$.
Answers and brief solutions
Show answers
- $5$ $c=\sqrt{16+9}=5$.
- Up and down The positive term contains $y^2$.
- $(-1,-1)$ and $(5,-1)$ Move $a=3$ horizontally from centre $(2,-1)$.
Synthesis and transfer
Differences in signal arrival time locate a transmitter on a hyperbola; multiple receiver pairs create intersecting curves that narrow the possible location.
For $x^2/16-y^2/9=1$, vertices lie at $(\pm4,0)$ and asymptotes are $y=\pm(3/4)x$. Since $c^2=a^2+b^2$, the foci are $(\pm5,0)$, outside the vertices. The difference—not the sum—of focal distances remains $2a=8$ on either branch. Far from the centre, the branches approach but never become the asymptote lines. Interchanging the positive and negative squared terms changes the opening direction, while translating the standard form moves every feature together. Substitution of a vertex and comparison with asymptote slope check the parameter reading.
Related topics
Teaching and accessibility note
Try it yourself
Hints are part of learning. Open one whenever it makes the next step feel possible.
For $x^2/16-y^2/9=1$, find $c$.
- $c=\sqrt{16+9}=5$.
End of lesson
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