Eccentricity and unified conic structure

Solution

Begin by identifying the mathematical object and the information that fixes it. Write the focus-directrix distance equation, square carefully, and compare the resulting standard form or compute $e$ from known a and $c$. The relevant conditions are not optional bookkeeping: Eccentricity is nonnegative and geometric. Sign changes in an equation do not make eccentricity negative. Following that structure gives It is an ellipse; set distance to focus equal to one-half distance to directrix and simplify.

Why this works

For standard ellipses and hyperbolas, $e=\frac{c}{a}$. Increasing $e$ stretches an ellipse toward a parabola and opens a hyperbola more strongly relative to its vertex distance. The calculation and the representation should agree, so a graph, diagram, table, or substitution check should support the same conclusion.

Worked example 1

A conic has focus at $(4,0),$ directrix $x=9,$ and eccentricity $\frac{1}{2}$. Determine its type and derive an equation.

Solution

Begin by identifying the mathematical object and the information that fixes it. Write the focus-directrix distance equation, square carefully, and compare the resulting standard form or compute $e$ from known a and $c$. The relevant conditions are not optional bookkeeping: Eccentricity is nonnegative and geometric. Sign changes in an equation do not make eccentricity negative. Following that structure gives It is an ellipse; set distance to focus equal to one-half distance to directrix and simplify.

Why this works

For standard ellipses and hyperbolas, $e=\frac{c}{a}$. Increasing $e$ stretches an ellipse toward a parabola and opens a hyperbola more strongly relative to its vertex distance. The calculation and the representation should agree, so a graph, diagram, table, or substitution check should support the same conclusion.

Problem

Relate $e=\frac{c}{a}$ for ellipses and hyperbolas.

Worked development

Write the focus-directrix distance equation, square carefully, and compare the resulting standard form or compute $e$ from known a and $c$. In this example, the first useful move is to make the defining structure visible rather than to search for a memorized answer. The ratio distinguishes conic families: $0<e<1$ gives an ellipse, $e=1$ a parabola, and $e>1$ a hyperbola. A circle is the limiting symmetric case $e=0$. Then apply the conditions explicitly: Eccentricity is nonnegative and geometric. Sign changes in an equation do not make eccentricity negative. Finish by checking the result in a second representation and explaining what the result means.

Interpretation

Eccentricity describes orbital shape, optical design, and a unified family of conics.

Problem

Show $e=1$ produces a parabola.

Worked development

Write the focus-directrix distance equation, square carefully, and compare the resulting standard form or compute $e$ from known a and $c$. In this example, the first useful move is to make the defining structure visible rather than to search for a memorized answer. The ratio distinguishes conic families: $0<e<1$ gives an ellipse, $e=1$ a parabola, and $e>1$ a hyperbola. A circle is the limiting symmetric case $e=0$. Then apply the conditions explicitly: Eccentricity is nonnegative and geometric. Sign changes in an equation do not make eccentricity negative. Finish by checking the result in a second representation and explaining what the result means.

Interpretation

Eccentricity describes orbital shape, optical design, and a unified family of conics.

Worked example 4: quick check

Classify a conic with eccentricity $1.4$.

Solution

Begin by identifying the mathematical object and the information that fixes it. Write the focus-directrix distance equation, square carefully, and compare the resulting standard form or compute $e$ from known a and $c$. The relevant conditions are not optional bookkeeping: Eccentricity is nonnegative and geometric. Sign changes in an equation do not make eccentricity negative. Following that structure gives Hyperbola.

Why this works

For standard ellipses and hyperbolas, $e=\frac{c}{a}$. Increasing $e$ stretches an ellipse toward a parabola and opens a hyperbola more strongly relative to its vertex distance. The calculation and the representation should agree, so a graph, diagram, table, or substitution check should support the same conclusion.

Connection forward

The next lesson solves intersections involving conic relations.