Ellipse
A circle stretched along one axis — the orbit of the planets, defined by two foci whose distances to any point always add to the same length
- The two-foci definition (constant sum \(2a\)) 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.
- How eccentricity \(e=\sqrt{1-\tfrac{b^2}{a^2}}\) sets the shape, and how the orientation depends on which denominator is larger.
- The latus rectum \(\tfrac{2b^2}{a}\) and the focal radii \(a\pm ex_1\).
- The parametric point \((a\cos\theta,b\sin\theta)\), the auxiliary circle, and the position test.
- Tangents and normals, the chord of contact, and the director circle of perpendicular tangents.
Definition of an Ellipse
Pin a loop of string around two tacks and trace it taut with a pencil: the curve is an ellipse. Each tack is a focus, and the construction enforces the defining property — the sum of the distances from any point on the curve to the two foci is constant, equal to \(2a\). Equivalently, an ellipse is the focus–directrix locus with eccentricity \(e<1\).
The Standard Equation
Centre the ellipse at the origin with foci at \((\pm ae,0)\) on the \(x\)-axis. Writing the constant-sum condition with the distance formula and simplifying gives the cleanest equation of the curve.
Centre \((0,0)\); vertices \((\pm a,0)\) and \((0,\pm b)\); major axis \(2a\) along \(x\), minor axis \(2b\); foci \((\pm ae,0)\); directrices \(x=\pm\tfrac{a}{e}\). The relation between the constants is \(b^2=a^2(1-e^2)\).
Eccentricity & Orientation
The eccentricity measures how far the ellipse departs from a circle. Which denominator is larger tells you along which axis the ellipse is stretched — and therefore where the foci sit.
When \(a>b\) the major axis lies along \(x\) and the foci are \((\pm ae,0)\). When \(b>a\) the major axis lies along \(y\) and the foci are \((0,\pm be)\). In both cases \(0
Before quoting any focus or directrix, decide which axis is major by comparing the denominators. The larger denominator is always \(a^2\) (the square of the semi-major axis); the foci always lie on the major axis, never the minor.
Latus Rectum & Focal Radii
The latus rectum is the focal chord perpendicular to the major axis. The focal radii are the two distances from a point on the ellipse to the foci — and the defining property makes their sum trivially \(2a\).
For a point \((x_1,y_1)\) on \(\tfrac{x^2}{a^2}+\tfrac{y^2}{b^2}=1\), the distances to the foci are \(a-ex_1\) and \(a+ex_1\), summing to \(2a\). The latus rectum has endpoints \(\left(\pm ae,\pm\tfrac{b^2}{a}\right)\).
Parametric Form & Position of a Point
The substitution \(x=a\cos\theta,\ y=b\sin\theta\) satisfies the equation automatically, with \(\theta\) the eccentric angle. Geometrically it comes from the auxiliary circle of radius \(a\) drawn on the major axis.
A point \((x_1,y_1)\) lies inside the ellipse when \(S_1<0\), on it when \(S_1=0\), and outside when \(S_1>0\).
Tangents
The same "split the squares" rule that worked for the circle and parabola gives the tangent at a point. As before there are three equivalent forms, and a slope tangent exists for every direction.
The first is the tangent at a point, the second at eccentric angle \(\theta\), the third the tangent of slope \(m\). A line \(y=mx+c\) is tangent precisely when \(c^2=a^2m^2+b^2\).
Normals & the Chord of Contact
The normal is perpendicular to the tangent at the point of contact. The chord of contact from an external point is again the equation \(T=0\), and the chord bisected at a point is \(T=S_1\).
The first is the normal at \((x_1,y_1)\); at eccentric angle \(\theta\) it reads \(ax\sec\theta-by\csc\theta=a^2-b^2\). The second, \(T=0\), gives the chord of contact from an external point — and the chord with midpoint \((x_1,y_1)\) is \(T=S_1\).
The Director Circle & Key Properties
Ask where two perpendicular tangents can meet, and the answer is a circle concentric with the ellipse — the director circle. A handful of further properties round out the chapter.
This is the locus of points from which a pair of perpendicular tangents can be drawn to the ellipse. Two other properties worth carrying: the reflection property (a ray from one focus reflects to the other), and that the auxiliary-circle foot \(Q\) of any ellipse point shares its \(x\)-coordinate.
Putting It to Work
Problem. Find the centre, foci, eccentricity and latus rectum of \(\dfrac{x^2}{25}+\dfrac{y^2}{9}=1\).
Solution. Here \(a^2=25,\ b^2=9\) with \(a>b\), so \(a=5,\ b=3\):
Problem. Find the equation of the ellipse with foci \((0,\pm3)\) and eccentricity \(\tfrac35\).
Solution. The foci are on the \(y\)-axis, so \(b>a\); from \(be=3\) and \(e=\tfrac35\) we get \(b=5\), then \(a^2=b^2(1-e^2)=16\):
Problem. Find the distances of the point \(\left(3,\tfrac{12}{5}\right)\) on \(\dfrac{x^2}{25}+\dfrac{y^2}{9}=1\) from the two foci.
Solution. With \(a=5,\ e=\tfrac45,\ x_1=3\), the focal radii are \(a\pm ex_1\):
Problem. Find the tangent to \(\dfrac{x^2}{25}+\dfrac{y^2}{9}=1\) at the point \(\left(3,\tfrac{12}{5}\right)\).
Solution. Apply \(\dfrac{xx_1}{a^2}+\dfrac{yy_1}{b^2}=1\):
Problem. Find the tangents to \(\dfrac{x^2}{16}+\dfrac{y^2}{9}=1\) with slope \(1\).
Solution. Use \(y=mx\pm\sqrt{a^2m^2+b^2}\) with \(a^2=16,\ b^2=9,\ m=1\):
Problem. Find the eccentricity and foci of \(9x^2+4y^2=36\).
Solution. Divide by \(36\) to get \(\dfrac{x^2}{4}+\dfrac{y^2}{9}=1\); here \(b^2=9>a^2=4\), so the major axis is along \(y\):
Chapter Summary
Sum of focal distances \(=2a\); eccentricity \(e<1\).
\(\tfrac{x^2}{a^2}+\tfrac{y^2}{b^2}=1\); foci \((\pm ae,0)\), directrices \(x=\pm\tfrac{a}{e}\).
\(b^2=a^2(1-e^2)\); major axis holds the foci.
Latus rectum \(\tfrac{2b^2}{a}\); focal radii \(a\pm ex_1\).
Point \((a\cos\theta,b\sin\theta)\); position from \(S_1\).
Tangent \(y=mx\pm\sqrt{a^2m^2+b^2}\); director circle \(x^2+y^2=a^2+b^2\).
Problems
Decide which axis is major first, read off \(a,b,e\), then apply the focal, tangent or normal result the question calls for. Difficulty rises down the list.
- Find the eccentricity, foci and vertices of \(\dfrac{x^2}{16}+\dfrac{y^2}{9}=1\).
- Find the lengths of the major and minor axes of \(\dfrac{x^2}{49}+\dfrac{y^2}{25}=1\).
- Find the equation of the ellipse with vertices \((\pm5,0)\) and foci \((\pm3,0)\).
- Find the length of the latus rectum of \(\dfrac{x^2}{36}+\dfrac{y^2}{16}=1\).
- Find the equation of the ellipse with foci \((0,\pm4)\) and eccentricity \(\tfrac45\).
- Find the eccentricity of \(4x^2+9y^2=36\).
- Find the point on \(\dfrac{x^2}{16}+\dfrac{y^2}{9}=1\) whose eccentric angle is \(60^{\circ}\).
- Find the tangent to \(\dfrac{x^2}{25}+\dfrac{y^2}{16}=1\) at the point \(\left(3,\tfrac{16}{5}\right)\).
- Find the tangents to \(\dfrac{x^2}{25}+\dfrac{y^2}{16}=1\) having slope \(1\).
- Determine whether the point \((1,2)\) lies inside or outside \(\dfrac{x^2}{9}+\dfrac{y^2}{4}=1\).
- Find the equation of the director circle of \(\dfrac{x^2}{16}+\dfrac{y^2}{9}=1\).
- Find the equation of the chord of \(\dfrac{x^2}{25}+\dfrac{y^2}{9}=1\) which is bisected at the point \((1,1)\).