Hyperbola
Two open branches racing toward a pair of asymptotes — the conic of constant distance-difference, and the close cousin of the ellipse with eccentricity past one
- The two-foci definition (constant difference \(2a\)) and the eccentricity \(e>1\).
- The standard equation \(\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1\) with its centre, vertices, foci, axes and directrices.
- The asymptotes \(y=\pm\tfrac{b}{a}x\) and the conjugate hyperbola that shares them.
- The latus rectum \(\tfrac{2b^2}{a}\) and the focal distances \(|ex_1\pm a|\).
- The parametric point \((a\sec\theta,b\tan\theta)\) and the position test.
- Tangents and normals, the chord of contact, and the rectangular hyperbola \(xy=c^2\).
Definition of a Hyperbola
The ellipse fixed the sum of two focal distances; the hyperbola fixes their difference. It is the locus of points whose distances to two fixed foci differ by a constant \(2a\). Holding the difference rather than the sum constant opens the curve into two branches, and the focus–directrix eccentricity is now \(e>1\).
The Standard Equation
Centre the hyperbola at the origin with foci at \((\pm ae,0)\). Writing the constant-difference condition and simplifying gives the standard equation — identical to the ellipse but with a minus sign.
Centre \((0,0)\); vertices \((\pm a,0)\); transverse axis \(2a\) (along \(x\)), conjugate axis \(2b\); foci \((\pm ae,0)\); directrices \(x=\pm\tfrac{a}{e}\). The relation between the constants is \(b^2=a^2(e^2-1)\) — note the plus sign in \(c=ae=\sqrt{a^2+b^2}\).
Asymptotes & the Conjugate Hyperbola
Far from the centre, a hyperbola hugs two straight lines through the centre — its asymptotes. Replacing the \(1\) on the right by \(0\) gives them at once; replacing it by \(-1\) gives the conjugate hyperbola, which opens the other way and shares the same asymptotes.
The asymptotes are the diagonals of the central rectangle of sides \(2a\) and \(2b\). The conjugate hyperbola has the same asymptotes; if the original has eccentricity \(e\) and the conjugate \(e'\), then \(\tfrac{1}{e^2}+\tfrac{1}{e'^2}=1\).
Latus Rectum & Focal Distances
The latus rectum keeps the same form as the ellipse. The focal distances of a point also mirror the ellipse — with the constant difference \(2a\) replacing the constant sum.
For a point \((x_1,y_1)\) on the right branch, the distances to the foci are \(ex_1-a\) and \(ex_1+a\), differing by \(2a\). The latus rectum has endpoints \(\left(\pm ae,\pm\tfrac{b^2}{a}\right)\).
Parametric Form & Position of a Point
The identity \(\sec^2\theta-\tan^2\theta=1\) matches the hyperbola exactly, giving its standard parametric point. Substituting into \(S\equiv\tfrac{x^2}{a^2}-\tfrac{y^2}{b^2}-1\) locates a point — but note the sign convention is opposite to the ellipse.
A point lies on the hyperbola when \(S_1=0\). Because the curve opens outward, a point on the focus side of a branch gives \(S_1>0\), while a point in the region between the branches gives \(S_1<0\).
Tangents
The "split the squares" rule again produces the tangent at a point, with the minus sign carried through. The slope tangent exists only for directions steeper than the asymptotes, which is exactly what the square root in the condition enforces.
The first is the tangent at a point, the second at parameter \(\theta\), the third the tangent of slope \(m\). A line \(y=mx+c\) is tangent precisely when \(c^2=a^2m^2-b^2\) — real only when \(m^2>\tfrac{b^2}{a^2}\), i.e. steeper than the asymptotes.
Normals & the Chord of Contact
The normal is perpendicular to the tangent at the point of contact; its equation carries a plus sign where the ellipse had a minus. The chord of contact from an external point is again \(T=0\), and the chord bisected at a point is \(T=S_1\).
The first is the normal at \((x_1,y_1)\). The second, \(T=0\), gives the chord of contact from an external point — and the chord with midpoint \((x_1,y_1)\) is \(T=S_1\), exactly as for the circle and ellipse.
The Rectangular Hyperbola
When the transverse and conjugate axes are equal, \(a=b\), the hyperbola is rectangular (equilateral): its asymptotes are perpendicular and its eccentricity is fixed. Rotating so the asymptotes become the coordinate axes gives the familiar form \(xy=c^2\).
Every rectangular hyperbola has eccentricity \(\sqrt2\), since \(e=\sqrt{1+\tfrac{b^2}{a^2}}=\sqrt2\) when \(a=b\). In the form \(xy=c^2\) the parametric point is \(\left(ct,\tfrac{c}{t}\right)\), and the tangent there is \(x+t^2y=2ct\).
Putting It to Work
Problem. Find the foci, eccentricity, latus rectum and asymptotes of \(\dfrac{x^2}{16}-\dfrac{y^2}{9}=1\).
Solution. Here \(a^2=16,\ b^2=9\), so \(a=4,\ b=3\):
Problem. Find the equation of the hyperbola with vertices \((\pm3,0)\) and a focus at \((5,0)\).
Solution. Here \(a=3,\ c=5\), and \(b^2=c^2-a^2=25-9=16\):
Problem. Find the distances of the point \((8,3\sqrt3)\) on \(\dfrac{x^2}{16}-\dfrac{y^2}{9}=1\) from the two foci.
Solution. With \(a=4,\ e=\tfrac54,\ x_1=8\), the focal distances are \(ex_1\mp a\):
Problem. Find the tangent to \(\dfrac{x^2}{16}-\dfrac{y^2}{9}=1\) at the point \((8,3\sqrt3)\).
Solution. Apply \(\dfrac{xx_1}{a^2}-\dfrac{yy_1}{b^2}=1\):
Problem. Find the tangents to \(\dfrac{x^2}{9}-\dfrac{y^2}{4}=1\) with slope \(1\).
Solution. Use \(y=mx\pm\sqrt{a^2m^2-b^2}\) with \(a^2=9,\ b^2=4,\ m=1\):
Problem. Find the eccentricity and asymptotes of \(16x^2-9y^2=144\).
Solution. Divide by \(144\) to get \(\dfrac{x^2}{9}-\dfrac{y^2}{16}=1\); here \(a=3,\ b=4\):
Chapter Summary
Difference of focal distances \(=2a\); eccentricity \(e>1\).
\(\tfrac{x^2}{a^2}-\tfrac{y^2}{b^2}=1\); foci \((\pm ae,0)\), \(b^2=a^2(e^2-1)\).
\(y=\pm\tfrac{b}{a}x\); conjugate hyperbola shares them, \(\tfrac1{e^2}+\tfrac1{e'^2}=1\).
Latus rectum \(\tfrac{2b^2}{a}\); focal distances \(|ex_1\pm a|\).
Tangent \(y=mx\pm\sqrt{a^2m^2-b^2}\); \(T=0\) and \(T=S_1\) as before.
\(a=b\Rightarrow e=\sqrt2\); \(xy=c^2\) with point \(\left(ct,\tfrac{c}{t}\right)\).
Problems
Read off \(a,b,e\) from the standard form, remembering the plus sign in \(c=\sqrt{a^2+b^2}\), then apply the focal, tangent or normal result. Difficulty rises down the list.
- Find the eccentricity, foci and vertices of \(\dfrac{x^2}{9}-\dfrac{y^2}{16}=1\).
- Find the asymptotes of \(\dfrac{x^2}{25}-\dfrac{y^2}{16}=1\).
- Find the equation of the hyperbola with vertices \((\pm2,0)\) and foci \((\pm3,0)\).
- Find the length of the latus rectum of \(\dfrac{x^2}{16}-\dfrac{y^2}{9}=1\).
- Find the equation of the hyperbola with foci \((0,\pm5)\) and vertices \((0,\pm3)\).
- Find the eccentricity of \(3x^2-2y^2=6\).
- Find the point on \(\dfrac{x^2}{16}-\dfrac{y^2}{9}=1\) whose parameter is \(\theta=45^{\circ}\).
- Find the tangent to \(\dfrac{x^2}{9}-\dfrac{y^2}{4}=1\) at the point \(\left(5,\tfrac83\right)\).
- Find the tangents to \(\dfrac{x^2}{16}-\dfrac{y^2}{9}=1\) having slope \(1\).
- Determine the position of the point \((4,1)\) relative to \(\dfrac{x^2}{9}-\dfrac{y^2}{4}=1\).
- Show that the eccentricity of a rectangular hyperbola is \(\sqrt2\).
- Find the equation of the chord of \(\dfrac{x^2}{16}-\dfrac{y^2}{9}=1\) which is bisected at the point \((5,3)\).