Ellipses in the plane

Ask the tutor about Ellipses in the plane Open the graphing workbench

A circle is one radius. Stretch one axis and the curve can still close. A hyperbola freezes a difference of distances and splits into two branches; a parabola freezes one focus against a directrix. The question is which freeze keeps a closed oval with two axes — and which extra placement then lets us write a Cartesian equation at all.

Area and circumference of an ellipse sit with circles, sectors, and ellipses. Cartesian axes sit with lines in the plane.

A constant sum, one closed curve

Fix two points $B_{-}$ and $B_{+}$ (the focal points). A point $P$ lies on this ellipse when the sum of distances is constant, equal to $2 a$ :

$P B_{-} + P B_{+} = 2 a .$

That is the string construction: pin a string of length $2 a$ at both foci, keep it taut with a pencil, and the pencil traces the oval. Swap the freeze from a sum to a difference and the locus cannot close — that is the hyperbola page. Drag $Q$ . The two segments change length; their sum does not. $B m$ and $B p$ stay put. The two segments on this board are that string.

The numerical eccentricity for this shape satisfies $0 \leq ε < 1$ . The ratio to a directrix that makes $ε$ visible lives on polar coordinates and conic sections. This page keeps the two-focus sum.

Drag Q. Bm is the focal point B−, Bp is B+. The two segments keep the same sum 2a.

Sit the center at the origin

The sum rule does not yet name coordinates. Extra assumptions for the rest of this page: the center sits at the origin, the long axis (length $a$ ) lies along $x$ , the short axis (length $b$ ) lies along $y$ , and $0 < b < a$ , so the figure is longer along $x$ and symmetric about the origin. Then the same set of points $(x, y)$ satisfies

$\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1.$

From $a$ and $b$ one also writes the linear eccentricity, the numerical eccentricity, and the half-parameter

$e = a^{2} - b^{2}, ε = \frac{e}{a}, p = \frac{b ^{2}}{a} .$

Linear eccentricity here is $a^{2} - b^{2}$ , not the hyperbola formula with a plus — that minus is why $ε < 1$ . The two points $(\pm e, 0)$ are $B_{\pm}$ ; on the first board they are $B m$ and $B p$ . The case $a = b$ collapses $e$ to $0$ : the two foci meet at the origin and the equation is a circle. Rotate the axes, or move the center off the origin, and $x^{2} / a^{2} + y^{2} / b^{2} = 1$ is the wrong formula for the same geometric freeze.

One $t$ walks the whole curve

The oval is closed, so one parameter can walk all of it. The same ellipse is traced by

$x = a cos t, y = b sin t, 0 \leq t < 2 π .$

This math handbook measures $t$ in radians. When $t$ runs from $0$ to $2 π$ , a point $Q$ on the ellipse travels once counter-clockwise. The starting value $t = 0$ is the vertex $(a, 0)$ on the positive $x$ -axis — the right-hand end of the long axis.

A hyperbola needs a sign in $x$ before $cosh t$ can start, and one formula never crosses to the other branch. Here there is no second branch to pick. Drop $0 \leq t < 2 π$ and you have not finished the oval; replace $cos$ and $sin$ by $cosh$ and $sinh$ and the curve will not close.

The tangent is that equation, linearized

At a point $(x_{0}, y_{0})$ already on $x^{2} / a^{2} + y^{2} / b^{2} = 1$ , the tangent line is

$\frac{x x _{0}}{a ^{2}} + \frac{y y _{0}}{b ^{2}} = 1.$

This is not a new curve. It is the Cartesian equation made linear at that point. Drag $P$ . The line through $P$ is that tangent. Leave the point, and the same $a$ and $b$ still name the same ellipse; only the line changes.

Drag P. The line through P is the tangent at that point — the Cartesian equation linearized there.

Leave one focus, meet the other

A further freeze: a light ray that leaves one focal point and reflects off the ellipse meets the other focal point. That is the same two points $B_{\pm}$ already on the first board, not a new pair. Drop “the ray started at a focus” and the bounce does not have to hit the other. This page has no live bounce; the hyperbola shows the two-focus reflection (backward extension), and the parabola shows the one-focus mirror (parallel to the axis).

The object is the sum freeze: two foci, constant $2 a$ , one closed curve. Under the extra placement (center at the origin, long axis along $x$ , $0 < b < a$ ) that freeze is the Cartesian equation $x^{2} / a^{2} + y^{2} / b^{2} = 1$ , with $B_{\pm}$ at $(\pm e, 0)$ and $e = a^{2} - b^{2}$ . The parameterization and the tangent are that equation along the oval and at a point.

Thaw $ε$ and the same family writes one polar equation: polar coordinates and conic sections.