An ellipse is the set of points the sum of whose distances from two fixed points (the foci) is constant. A hyperbola is the set of points the difference of whose distances from two foci is constant. One word changes — sum versus difference — and a closed oval becomes a pair of opening branches. Both are conics with a focus–directrix description $PF = e\cdot PM$, the ellipse with $0 < e < 1$ and the hyperbola with $e > 1$.
Ellipse (foci on the $x$-axis, $a > b$):
$$\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1, \qquad a > b > 0$$
Here $2a$ is the major axis (along the $x$-axis) and $2b$ the minor axis. The vertices (ends of the major axis) are $(\pm a, 0)$ and the ends of the minor axis are $(0, \pm b)$. The foci sit at $(\pm c, 0)$ where:
$$c = \sqrt{a^2 - b^2}, \qquad \text{equivalently } b^2 = a^2(1 - e^2)$$
The defining "constant sum" equals $2a$ — the full length of the major axis. If instead the larger denominator sits under $y^2$, the major axis is vertical: the roles of $a$ and $b$ swap and the foci move to $(0, \pm c)$.
Hyperbola (foci on the $x$-axis):
$$\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$$
The vertices are $(\pm a, 0)$, the foci are $(\pm c, 0)$, and the two branches approach the slanted asymptotes $y = \pm \dfrac{b}{a}x$. The relation is now a sum under the root:
$$c = \sqrt{a^2 + b^2}, \qquad \text{equivalently } b^2 = a^2(e^2 - 1)$$
For both curves the shape is captured by the eccentricity, and both share the same latus-rectum formula:
$$e = \dfrac{c}{a}, \qquad \text{Length of latus rectum} = \dfrac{2b^2}{a}$$
The lone difference in the $c$-relations — a minus for the ellipse, a plus for the hyperbola — is exactly what pushes the eccentricity below $1$ in one case and above $1$ in the other. Finding the equation from conditions is a matter of extracting $a$, $b$ and $c$ from the data: vertices give $a$, foci give $c$, eccentricity gives the ratio $c/a$, and then the appropriate relation supplies the missing one. Word problems (the arch of a bridge, a whispering gallery, the orbit of a planet) usually reduce to fitting an ellipse or hyperbola to a few measured points.
Deeper Insight — eccentricity is the dial that turns one conic into another: Every conic in this chapter is a slice of a double cone, and a single number, the eccentricity $e$, records how steeply the slicing plane is tilted. A circle is the perfectly level cut with $e = 0$; tilt a little and you get an ellipse with $0 < e < 1$; tilt until the plane runs parallel to the cone's side and the closed oval breaks open into a parabola with $e = 1$; tilt further still and you cut both nappes of the cone, producing a hyperbola with $e > 1$. So ellipses, parabolas and hyperbolas are not three unrelated curves but three settings of the same dial — which is why $e$ appears in the single focus–directrix law $PF = e\cdot PM$ for all of them. The sign flip in $c = \sqrt{a^2 \mp b^2}$ is the algebraic shadow of this geometry: for an ellipse the foci sit inside, so $c < a$ and $e < 1$; for a hyperbola the foci lie beyond the vertices, so $c > a$ and $e > 1$. Hold on to the cone picture and the whole chapter reads as one story rather than a list of formulae.
Find the foci, vertices and eccentricity of the ellipse $\dfrac{x^2}{25} + \dfrac{y^2}{9} = 1$.
Solution- Here $a^2 = 25$, $b^2 = 9$, so $a = 5$, $b = 3$ (major axis along the $x$-axis).
- $c = \sqrt{a^2 - b^2} = \sqrt{25 - 9} = \sqrt{16} = 4$.
- Vertices: $(\pm 5, 0)$; foci: $(\pm 4, 0)$.
- Eccentricity $e = \dfrac{c}{a} = \dfrac{4}{5}$.
Answer: Vertices $(\pm 5, 0)$, foci $(\pm 4, 0)$, $e = \dfrac{4}{5}$.
Find the foci, vertices and eccentricity of the hyperbola $\dfrac{x^2}{16} - \dfrac{y^2}{9} = 1$.
Solution- Here $a^2 = 16$, $b^2 = 9$, so $a = 4$, $b = 3$.
- For a hyperbola $c = \sqrt{a^2 + b^2} = \sqrt{16 + 9} = \sqrt{25} = 5$.
- Vertices: $(\pm 4, 0)$; foci: $(\pm 5, 0)$.
- Eccentricity $e = \dfrac{c}{a} = \dfrac{5}{4}$.
Answer: Vertices $(\pm 4, 0)$, foci $(\pm 5, 0)$, $e = \dfrac{5}{4}$.
Find the length of the latus rectum of the ellipse $\dfrac{x^2}{36} + \dfrac{y^2}{20} = 1$.
Solution- Here $a^2 = 36$, $b^2 = 20$, so $a = 6$.
- Length of latus rectum $= \dfrac{2b^2}{a} = \dfrac{2(20)}{6} = \dfrac{40}{6} = \dfrac{20}{3}$.
Answer: $\dfrac{20}{3}$.
Find the equation of the ellipse with vertices $(\pm 6, 0)$ and foci $(\pm 4, 0)$.
Solution- Vertices $(\pm a, 0)$ give $a = 6$, so $a^2 = 36$.
- Foci $(\pm c, 0)$ give $c = 4$, and $b^2 = a^2 - c^2 = 36 - 16 = 20$.
- The equation is $\dfrac{x^2}{36} + \dfrac{y^2}{20} = 1$.
Answer: $\dfrac{x^2}{36} + \dfrac{y^2}{20} = 1$.
Find the equations of the asymptotes of the hyperbola $\dfrac{x^2}{9} - \dfrac{y^2}{4} = 1$.
Solution- Here $a^2 = 9$, $b^2 = 4$, so $a = 3$, $b = 2$.
- The asymptotes of $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$ are $y = \pm \dfrac{b}{a}x$.
- Substitute: $y = \pm \dfrac{2}{3}x$.
Answer: $y = \dfrac{2}{3}x$ and $y = -\dfrac{2}{3}x$.
Find the equation of the hyperbola with foci $(\pm 5, 0)$ and eccentricity $\dfrac{5}{3}$.
Solution- Foci $(\pm c, 0)$ give $c = 5$; eccentricity $e = \dfrac{c}{a} = \dfrac{5}{3}$ gives $a = \dfrac{c}{e} = \dfrac{5}{5/3} = 3$, so $a^2 = 9$.
- For a hyperbola $b^2 = c^2 - a^2 = 25 - 9 = 16$.
- The equation is $\dfrac{x^2}{9} - \dfrac{y^2}{16} = 1$.
Answer: $\dfrac{x^2}{9} - \dfrac{y^2}{16} = 1$.
Find the eccentricity, foci and lengths of both axes of the ellipse $\dfrac{x^2}{16} + \dfrac{y^2}{9} = 1$.
Solution- $a^2 = 16$, $b^2 = 9$, so $a = 4$, $b = 3$ (major axis along the $x$-axis).
- $c = \sqrt{a^2 - b^2} = \sqrt{16 - 9} = \sqrt{7}$.
- Eccentricity $e = \dfrac{c}{a} = \dfrac{\sqrt{7}}{4}$; foci $(\pm \sqrt{7}, 0)$.
- Major axis $= 2a = 8$, minor axis $= 2b = 6$.
Answer: $e = \dfrac{\sqrt{7}}{4}$, foci $(\pm \sqrt{7}, 0)$, major axis $8$, minor axis $6$.
Find the equation of the ellipse whose major axis is along the $x$-axis, that passes through $(4, 3)$ and $(6, 2)$.
Solution- Take the form $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$ and let $\dfrac{1}{a^2} = p$, $\dfrac{1}{b^2} = q$.
- Through $(4,3)$: $16p + 9q = 1$. Through $(6,2)$: $36p + 4q = 1$.
- Solve: from the two equations, $p = \dfrac{1}{52}$ and $q = \dfrac{1}{13}$, so $a^2 = 52$, $b^2 = 13$.
- Hence $\dfrac{x^2}{52} + \dfrac{y^2}{13} = 1$.
Answer: $\dfrac{x^2}{52} + \dfrac{y^2}{13} = 1$.
Find the eccentricity, foci and length of the latus rectum of the hyperbola $9x^2 - 16y^2 = 144$.
Solution- Divide through by $144$ to reach standard form: $\dfrac{x^2}{16} - \dfrac{y^2}{9} = 1$.
- So $a^2 = 16$, $b^2 = 9$, giving $a = 4$, $b = 3$.
- $c = \sqrt{a^2 + b^2} = \sqrt{25} = 5$, so $e = \dfrac{c}{a} = \dfrac{5}{4}$ and the foci are $(\pm 5, 0)$.
- Latus rectum $= \dfrac{2b^2}{a} = \dfrac{2(9)}{4} = \dfrac{9}{2}$.
Answer: $e = \dfrac{5}{4}$, foci $(\pm 5, 0)$, latus rectum $\dfrac{9}{2}$.
A rod $12$ m long moves with its ends on two perpendicular rails (the coordinate axes). Find the locus of a point $P$ on the rod that is $3$ m from the end on the $x$-axis.
Solution- Let the rod meet the $x$-axis at $A(\alpha, 0)$ and the $y$-axis at $B(0, \beta)$, with $\alpha^2 + \beta^2 = 12^2 = 144$.
- $P$ divides $AB$ so that $PA = 3$ and $PB = 9$; using the section formula, $P = (x, y)$ with $x = \dfrac{9\alpha}{12} = \dfrac{3\alpha}{4}$ and $y = \dfrac{3\beta}{12} = \dfrac{\beta}{4}$.
- So $\alpha = \dfrac{4x}{3}$ and $\beta = 4y$; substitute into $\alpha^2 + \beta^2 = 144$: $\dfrac{16x^2}{9} + 16y^2 = 144$.
- Divide by $144$: $\dfrac{x^2}{81} + \dfrac{y^2}{9} = 1$.
Answer: The locus is the ellipse $\dfrac{x^2}{81} + \dfrac{y^2}{9} = 1$.
A semi-elliptical archway over a road has a span (width) of $8$ m and a maximum height of $2$ m at the centre. Find the height of the arch $1.5$ m from the centre.
Solution- Model the arch as the top half of $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$ with the centre of the span at the origin.
- Half-span $= 4$ m gives $a = 4$; maximum height $= 2$ m gives $b = 2$. So $\dfrac{x^2}{16} + \dfrac{y^2}{4} = 1$.
- At $x = 1.5$: $\dfrac{(1.5)^2}{16} + \dfrac{y^2}{4} = 1 \Rightarrow \dfrac{2.25}{16} + \dfrac{y^2}{4} = 1 \Rightarrow \dfrac{y^2}{4} = 1 - 0.140625 = 0.859375$.
- So $y^2 = 3.4375 \Rightarrow y \approx 1.85$ m.
Answer: The arch is about $1.85$ m high at $1.5$ m from the centre.
Find the equation of the ellipse with foci $(0, \pm 4)$ and eccentricity $\dfrac{4}{5}$ (major axis vertical).
Solution- Foci on the $y$-axis mean the major axis is vertical: $\dfrac{x^2}{b^2} + \dfrac{y^2}{a^2} = 1$ with $a > b$, foci $(0, \pm c)$.
- $c = 4$ and $e = \dfrac{c}{a} = \dfrac{4}{5}$ give $a = \dfrac{c}{e} = \dfrac{4}{4/5} = 5$, so $a^2 = 25$.
- $b^2 = a^2 - c^2 = 25 - 16 = 9$.
- Hence $\dfrac{x^2}{9} + \dfrac{y^2}{25} = 1$.
Answer: $\dfrac{x^2}{9} + \dfrac{y^2}{25} = 1$.