Conic Sections: Chapter Review

Vocabulary and notation

Focus, directrix

The fixed point and the fixed line of a locus condition. A parabola has one of each; this chapter defines an ellipse and a hyperbola with two foci instead, so no directrix is used there. A directrix answer is an equation such as $y = -3$, never a point.

Axis, vertex, center

A parabola's axis runs from the focus perpendicular to the directrix, its vertex the midpoint between them. An ellipse or hyperbola is centered at the midpoint of its foci.

Major axis, minor axis, co-vertices

Ellipse only. The major axis runs through the foci, length $2a$; the minor axis is perpendicular through the center, length $2b$, and its endpoints are the co-vertices.

Transverse axis, conjugate axis

Hyperbola only. The transverse axis joins the vertices, length $2a$. The conjugate axis is perpendicular through the center, length $2b$, and touches no point of the curve.

Focal width

The chord through a focus parallel to the directrix. For a parabola it is $|4p|$, so from the focus step $|2p|$ each way for two more points.

Central rectangle

Hyperbola: the rectangle about the center reaching $a$ along the transverse axis and $b$ along the conjugate axis. Its diagonals are the asymptotes, and center to corner is exactly $c$.

Degenerate case

A collapsed locus: the focal sum shrinking to $2a = 2c$ flattens an ellipse to the segment joining its foci (a different limit, both axes shrinking to $0$ together, collapses it to a single point instead), $2a \ge 2c$ collapses a hyperbola to two rays or nothing, and a parabola collapses to a line when its focus sits on its directrix.

Formulas and theorems

• The four distance conditions

  $$\begin{gathered} \text{circle:}\ PC = r \\ \text{parabola:}\ PF = \operatorname{dist}(P, d) \\ \text{ellipse:}\ PF_1 + PF_2 = 2a \\ \text{hyperbola:}\ \bigl| PF_1 - PF_2 \bigr| = 2a \end{gathered}$$

  Use when In order: circle, parabola, ellipse, hyperbola, with $2c$ the focus separation. Needs $r > 0$; the focus off $d$; $a > c \ge 0$ (ellipse); $0 < a < c$ (hyperbola). The absolute value admits both branches. Measure with $PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$, and to a line along the perpendicular: $|y_0 - m|$ to the horizontal $y = m$, $|x_0 - m|$ to the vertical $x = m$. A midpoint averages endpoints, so the ends of a diameter give the center, and HALF their distance is the radius.

• Circle: center-radius form, general form, and the sign that decides

  $$\begin{gathered} (x - h)^2 + (y - k)^2 = r^2 \\ \Longleftrightarrow\ x^2 + y^2 + Dx + Ey + F = 0 \\ \bigl(x + \tfrac{D}{2}\bigr)^2 + \bigl(y + \tfrac{E}{2}\bigr)^2 \\ = \tfrac{D^2 + E^2 - 4F}{4} \end{gathered}$$

  Use when The form SUBTRACTS, and the right side is $r^2$, not $r$. Needs equal, nonzero coefficients on $x^2$ and $y^2$ and no $xy$ term; divide by that coefficient first. Center $\left(-\tfrac{D}{2}, -\tfrac{E}{2}\right)$ whatever the signs. The sign of $D^2 + E^2 - 4F$ alone settles it: positive a circle of radius half its square root, zero that center alone, negative nothing real.

  e.g. $2x^2 + 2y^2 - 12x + 4y - 6 = 0$: divide by $2$, then $(x - 3)^2 + (y + 1)^2 = 13$.

• Parabola: standard forms, and the bridge to a quadratic

  $$\begin{gathered} (x - h)^2 = 4p\,(y - k) \\ (y - k)^2 = 4p\,(x - h) \\ y = a(x - h)^2 + k \ \Longrightarrow\ 4p = \tfrac{1}{a} \end{gathered}$$

  Text description

  An upward opening parabola with its vertex on the axis, its focus a distance p above the vertex, and its directrix the same distance p below.

  Use when Vertex $(h, k)$; $p \neq 0$ is the SIGNED vertex-to-focus distance, so the number in front is $4p$, never $p$, and the square must stand ALONE before reading it. Vertical: focus $(h, k + p)$, directrix $y = k - p$. Horizontal: focus $(h + p, k)$, directrix $x = h - p$, and not a function of $x$. A positive $p$ opens the curve up or right, a negative one down or left, always bending toward the focus; a point's focal distance equals its distance to the directrix, and every ray parallel to the axis reflects through the focus.

  e.g. $(x + 2)^2 = -8(y - 1)$: $p = -2$, focus $(-2, -1)$, directrix $y = 3$.

• Ellipse: standard form

  $$\begin{gathered} \frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1 \quad (\text{wide}) \\ \frac{(x - h)^2}{b^2} + \frac{(y - k)^2}{a^2} = 1 \quad (\text{tall}) \\ c^2 = a^2 - b^2 \end{gathered}$$

  Text description

  An ellipse with center, semi-major axis a, semi-minor axis b, and a focus c from the center; legs b and c meet at a right angle with hypotenuse a.

  Use when Center $(h, k)$, right side exactly $1$, $a \ge b > 0$, with $a = b$ the circle. $a^2$ is the LARGER denominator, never the one written first, and the major axis follows that variable's axis. Vertices $a$ from the center, co-vertices $b$, foci $c$, foci always on the MAJOR axis. Two focal distances total $2a$, each over $[a - c,\ a + c]$, extremes at the vertices.

  e.g. $\frac{x^2}{9} + \frac{y^2}{25} = 1$ is tall: vertices $(0, \pm 5)$, co-vertices $(\pm 3, 0)$, foci $(0, \pm 4)$.

• Hyperbola: standard form

  $$\begin{gathered} \frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1 \quad (\text{left-right}) \\ \frac{(y - k)^2}{a^2} - \frac{(x - h)^2}{b^2} = 1 \quad (\text{up-down}) \\ c^2 = a^2 + b^2 \end{gathered}$$

  Text description

  A hyperbola opening left and right, with its center, a vertex a from the center, and a focus c from the center lying beyond that vertex.

  Use when Center $(h, k)$, $a, b > 0$, NO size relation between them. $a^2$ sits under the POSITIVE term, whichever variable carries it, and that sign alone fixes the orientation: the first opens left and right, the second up and down. Vertices sit $a$ from the center and foci $c$, both along that positive term's axis. Since $c > a$ always, each focus lies beyond its vertex, and the strip between the vertices is empty.

  e.g. $\frac{y^2}{4} - \frac{x^2}{9} = 1$ opens up and down, vertices $(0, \pm 2)$, foci $(0, \pm\sqrt{13})$, though $9 > 4$.

• Hyperbola: asymptotes, by replacing $1$ with $0$

  $$\begin{gathered} \frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 0 \\ \Longrightarrow\ y - k = \pm\tfrac{b}{a}(x - h) \\ \text{gap}(x) = \frac{ab}{x + \sqrt{x^2 - a^2}} \end{gathered}$$

  Text description

  A hyperbola opening left and right, with the central rectangle of half-width a and half-height b whose extended diagonals are the asymptotes the branches approach.

  Use when Slopes $\pm\frac{b}{a}$ when $x^2$ is the positive term, $\pm\frac{a}{b}$ when $y^2$ is, so set the right side to $0$ rather than recall which. Both lines cross the center along the central rectangle's diagonals. The gap from $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ down to $y = \frac{b}{a}x$ ($x \ge a$) is positive at every such $x$ and never above $\frac{ab}{x}$, so a branch never arrives.

  e.g. $\frac{y^2}{9} - \frac{x^2}{16} = 1$ has slopes $\pm\frac{3}{4}$, not $\pm\frac{4}{3}$.

• Eccentricity separates ellipse from hyperbola

  $$\begin{gathered} e = \frac{c}{a} \\ \text{ellipse:}\ 0 \le e < 1 \\ \text{hyperbola:}\ e > 1 \end{gathered}$$

  Use when Ellipse and hyperbola only; a parabola has no $a$ or $c$ in this chapter's sense, so this ratio is not defined for one here. $a > 0$. Ellipse: $\frac{b}{a} = \sqrt{1 - e^2}$, and $e = 0$ is exactly the circle, foci merged. Hyperbola: $\frac{b}{a} = \sqrt{e^2 - 1}$, the asymptote slope when $x^2$ is positive, its reciprocal when $y^2$ is. Rectangular hyperbola: $a = b$, perpendicular asymptotes of slope $\pm 1$, $e = \sqrt{2}$.

  e.g. $\frac{x^2}{9} - \frac{y^2}{16} = 1$ has $c = 5$ and $e = \tfrac{5}{3}$.

• How many points two circles share

  $$\begin{gathered} \text{two points:}\ |r_1 - r_2| < d < r_1 + r_2 \\ \text{one point:}\ d = r_1 + r_2 \\ \text{or}\ d = |r_1 - r_2| \end{gathered}$$

  Use when $d$ is the center separation, the circles distinct, and internal tangency also needs $r_1 \neq r_2$. Outside those ranges they share nothing, missing for opposite reasons: too far apart, or one buried inside the other. When the centers differ, subtracting the general forms cancels $x^2 + y^2$ and leaves the LINE through both intersection points. Concentric circles ($d = 0$) subtract to a false constant instead, confirming what the ranges above already say: distinct concentric circles share no points.

  e.g. Centers $(1, 2)$ and $(7, 10)$ with radii $2$ and $6$: $d = 10 > 8$, so they share nothing.

Problem types, step by step

Exam traps

• Trap Balancing a completed square with a plus when its bracket carries a negative coefficient: $4(x^2 + 2x) - 9(y^2 - 4y) = 68$ finished as $68 + 4 + 36 = 108$.

  Fix That bracket is multiplied by $-9$, so adding $4$ inside adds $-36$: the right side is $68 + 4 - 36 = 36$, giving $\frac{(x + 1)^2}{9} - \frac{(y - 2)^2}{4} = 1$.

• Trap Reading the center off the printed signs: $(x + 5)^2 + (y - 2)^2 = 9$ recorded as center $(5, -2)$.

  Fix The form subtracts, so $x + 5$ is $x - (-5)$: the center is $(-5, 2)$, the values making each squared term zero.

• Trap Running one focal relation for both curves, so an ellipse with $a = 5$ and $b = 3$ is given $c = \sqrt{34}$.

  Fix An ellipse subtracts, $c^2 = a^2 - b^2 = 16$, so $c = 4$; only a hyperbola adds. A negative $b^2$, or foci outside an ellipse, flags the wrong one.

• Trap Calling the FIRST denominator $a^2$: $\frac{x^2}{16} + \frac{y^2}{25} = 1$ read as a wide ellipse with vertices $(\pm 4, 0)$.

  Fix $a^2$ is the larger denominator, $25$, sitting under $y$, so the ellipse is tall: vertices $(0, \pm 5)$, foci $(0, \pm 3)$.

• Trap Carrying that larger-denominator rule to a hyperbola, so $\frac{y^2}{16} - \frac{x^2}{81} = 1$ is read as opening left and right.

  Fix The positive term names the axis and holds $a^2$, so it opens up and down, vertices $(0, \pm 4)$. Nothing requires $a > b$ here.

• Trap Quoting slopes $\pm\frac{b}{a}$ for a vertically opening hyperbola: $\frac{y^2}{25} - \frac{x^2}{4} = 1$ given $\pm\frac{2}{5}$.

  Fix Replace the $1$ with $0$: $\frac{y^2}{25} = \frac{x^2}{4}$ gives $y = \pm\frac{5}{2}x$.

• Trap Taking the number in front of a parabola's square as $p$: $y = 3(x + 1)^2 - 4$ given $p = 3$.

  Fix Isolate the square first: $(x + 1)^2 = \tfrac{1}{3}(y + 4)$, so $4p = \tfrac{1}{3}$ and $p = \tfrac{1}{12}$.