← M408MWeek 4: Conic Sections
M408M • Week 4Stewart §10.5

Conic Sections

Learn how distances define conic sections.
Use equations to find key points and sketch each curve.

Parabolas · Ellipses · Hyperbolas

1 Understand the definitions
2 Find the key points
3 Sketch the curve
Geometric definitions

Definitions of the three conics

Intersecting a cone with a plane produces a conic. We can also define each conic using distances.

Parabola: equal distances

$$PF=d(P,\ell)$$

Every point on a parabola is equally far from a fixed point (the focus) and a fixed line (the directrix).

Ellipse: fixed sum

$$PF_1+PF_2=2a$$

Add the distances from a point on the ellipse to the two foci. The total is the same everywhere on the ellipse.

Hyperbola: fixed difference

$$|PF_1-PF_2|=2a$$

Subtract the shorter distance to a focus from the longer one. The difference is the same everywhere on the hyperbola.

Parabola

Parabola: equal distances to a focus and directrix

Press Play to move P along the parabola. Its distance to the focus equals its perpendicular distance to the directrix.

Vertex at the origin

$$x^2=4py,\qquad p\ne0$$

Focus $(0,p)$ · directrix $y=-p$
Vertex $(0,0)$ · axis $x=0$

The vertex is halfway between the focus and directrix. $p>0$ opens upward; $p<0$ opens downward.

Derive the equation from equal distances
$$\sqrt{x^2+(y-p)^2}=|y+p|$$
$$x^2+(y-p)^2=(y+p)^2$$
$$x^2=4py$$
Swap $x$ and $y$: $y^2=4px$ opens right for $p>0$ and left for $p<0$.

The control excludes p = 0, which gives the line x = 0 instead of a parabola.

Ellipse • Distance definition

Ellipse: the distances to the foci add to 2a

Press Play to move P along the ellipse. The two focal distances change, but their sum always equals 2a.

Center at the origin

$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,\qquad a>b>0$$

Foci $(\pm c,0)$ · vertices $(\pm a,0)$
Center $(0,0)$ · major axis $y=0$

The semi-major axis is $a$, semi-minor axis is $b$, and $c^2=a^2-b^2$ ($c<a$). Distance sum: $PF_1+PF_2=2a$.

Derive the equation from distance sum
$$\sqrt{(x+c)^2+y^2}+\sqrt{(x-c)^2+y^2}=2a$$
$$\sqrt{(x-c)^2+y^2}=2a-\sqrt{(x+c)^2+y^2}$$
$$a\sqrt{(x+c)^2+y^2}=a^2+cx$$
$$(a^2-c^2)x^2+a^2y^2=a^2(a^2-c^2)$$
$$\text{Let } b^2=a^2-c^2\implies\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$
Swap $x$ and $y$: $\frac{x^2}{b^2}+\frac{y^2}{a^2}=1$ ($a>b$) has a vertical major axis. The diagram illustrates $a=5, b=3$ ($c=4$).
Ellipse • Why the formula works

Use a right triangle to find c

The top point is equally far from both foci. Since the two distances add to 2a, each distance is a.

$$a^2=b^2+c^2\quad\Longrightarrow\quad c^2=a^2-b^2$$

$c$ measures the distance from the center to either focus.

For $a=5$ and $b=3$, $c=\sqrt{25-9}=4$.

The foci lie inside the ellipse: $c<a$. If $a=b$, then $c=0$ and the ellipse is a circle.
The right triangle has legs b and c and hypotenuse a.
Ellipse • Orientation

Find the ellipse’s major axis

First divide so the right-hand side equals 1. The larger denominator identifies the major axis.

$$\frac{x^2}{9}+\frac{y^2}{25}=1$$

The larger denominator is under $y^2$, so the major axis is vertical.

$a=5$, $b=3$, $c=4$.

  • Vertices: $(0,\pm5)$
  • Foci: $(0,\pm4)$
  • Minor-axis endpoints: $(\pm3,0)$
$a$ is half the major-axis length. Its square goes under $x^2$ for a horizontal major axis and under $y^2$ for a vertical one.
The major axis is vertical.
Blue: vertices · orange: foci · teal: minor-axis endpoints.
Hyperbola • Distance definition

Hyperbola: the distance difference is 2a

Press Play to move P along either branch. The difference between distances to the foci always equals 2a.

Center at the origin

$$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1,\qquad a>0,\ b>0$$

Foci $(\pm c,0)$ · vertices $(\pm a,0)$
Center $(0,0)$ · transverse axis along $y=0$

Vertices are at distance $a$ from center, and $c^2=a^2+b^2$ ($c>a$). Distance difference: $|PF_1-PF_2|=2a$.

Derive the equation from distance difference
$$\left|\sqrt{(x+c)^2+y^2}-\sqrt{(x-c)^2+y^2}\right|=2a$$
$$\sqrt{(x+c)^2+y^2}=\pm2a+\sqrt{(x-c)^2+y^2}$$
$$cx-a^2=\pm a\sqrt{(x-c)^2+y^2}$$
$$(c^2-a^2)x^2-a^2y^2=a^2(c^2-a^2)$$
$$\text{Let } b^2=c^2-a^2\implies\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$$
Swap terms: $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$ opens vertically with vertices $(0,\pm a)$. The diagram illustrates $a=3, b=4$ ($c=5$).
Hyperbola • Why the formula works

Use a right triangle to find c

The corner of the guide rectangle is at (a, b). Its distance from the center is the focal distance c.

$$c^2=a^2+b^2\quad\Longrightarrow\quad c=\sqrt{a^2+b^2}$$

$c$ is the hypotenuse of a right triangle with legs $a$ and $b$.

For $a=3$ and $b=4$, $c=\sqrt{9+16}=\sqrt{25}=5$.

The foci lie outside the hyperbola: $c>a$. An arc of radius $c$ centered at the origin connects the rectangle corner $(a,b)$ directly to the focus $(c,0)$.
The right triangle has legs a and b and hypotenuse c.
Dashed orange arc: radius c connects corner (a, b) to focus F.
Hyperbola • Sketching

Find a hyperbola’s vertices and asymptotes

The positive squared term gives the opening direction. Use a rectangle to find the asymptote slopes.

$$\frac{x^2}{9}-\frac{y^2}{16}=1$$

Horizontal: vertices (±3, 0); asymptotes y = ±(4/3)x.

  1. Mark the center and the vertices.
  2. Draw a rectangle centered at the origin: 3 units left and right, and 4 units up and down. For the vertical example, swap 3 and 4.
  3. Extend the rectangle’s diagonals to draw the asymptotes.
  4. Draw one branch from each vertex, approaching the asymptotes as it moves away from the center.
$a^2$ is the denominator of the positive squared term. It need not be the larger denominator.
Dashed teal: asymptotes. Gray: guide rectangle.
Rectangle corners are not points on the curve.
Shifted conics

Shift a conic horizontally and vertically

Replacing x with x − h and y with y − k moves every point by h horizontally and k vertically.

$$\frac{(x-h)^2}{25}+\frac{(y-k)^2}{9}=1$$
  • Center: $(h,k)$
  • Vertices: $(h\pm5,k)$
  • Foci: $(h\pm4,k)$
  • Minor-axis endpoints: $(h,k\pm3)$
For example, $(x+2)^2$ means $h=-2$: a shift 2 units left. The lengths $a$, $b$, and $c$ stay the same.

For a shifted parabola, $(x-h)^2=4p(y-k)$: $(h,k)$ is the vertex, the focus is $(h,k+p)$, and the directrix is $y=k-p$.

Center(-2.00, 1.00)
Vertices(-7.00, 1.00)(3.00, 1.00)
Foci(-6.00, 1.00)(2.00, 1.00)
Exercises • Three Conics

Exercises: three conic sections

For each conic: complete the square into standard form, find all key geometric features, and sketch the curve.

Exercise 1: Parabola

$$x^2+4x+8y-12=0$$

Exercise 2: Ellipse

$$4x^2+9y^2-8x+36y+4=0$$

Exercise 3: Hyperbola

$$4y^2-9x^2-8y-36x-68=0$$
Exercise • Parabola

Exercise: a parabola

Complete the square to find the vertex, opening direction, focus, and directrix.

$$x^2+4x+8y-12=0$$

1. Isolate the squared variable:

$$x^2+4x=-8y+12$$

2. Complete the square on $x$:

$$(x+2)^2=-8y+12+4=-8y+16$$

3. Factor into standard form $(x-h)^2=4p(y-k)$:

$$\boxed{(x+2)^2=-8(y-2)}$$
The vertex $(-2,2)$ is halfway between the focus $(-2,0)$ and directrix $y=4$.
Blue: vertex $(-2,2)$ · orange: focus $(-2,0)$ · teal: directrix $y=4$ · dashed: axis $x=-2$.
Key features
  • Vertex: $(h,k)=(-2,2)$
  • $4p=-8\implies p=-2<0$ (opens downward)
  • Focus: $(h,k+p)=(-2,2-2)=(-2,0)$
  • Directrix: $y=k-p=2-(-2)\implies y=4$
  • Axis of symmetry: $x=-2$
Exercise • Ellipse

Exercise: an ellipse

Complete both squares to find the center, orientation, vertices, foci, and minor-axis endpoints.

$$4x^2+9y^2-8x+36y+4=0$$

1. Group terms and factor coefficients:

$$4(x^2-2x)+9(y^2+4y)=-4$$

2. Complete both squares and balance the RHS:

$$4(x^2-2x+1)+9(y^2+4y+4)=-4+4+36=36$$

3. Divide by 36 to get standard form:

$$\boxed{\frac{(x-1)^2}{9}+\frac{(y+2)^2}{4}=1}$$
Multiply before adding to RHS: $4\times1=4$ and $9\times4=36$. RHS: $-4+4+36=36$.
Blue: vertices $(-2,-2),(4,-2)$ · orange: foci $(1\pm\sqrt{5},-2)$ · teal: minor endpoints $(1,0),(1,-4)$.
Key features
  • Center: $(1,-2)$ · horizontal major axis ($9>4\implies a=3, b=2$)
  • $c=\sqrt{a^2-b^2}=\sqrt{9-4}=\sqrt{5}\approx2.24$
  • Vertices: $(1\pm3,-2)\implies(-2,-2)\text{ and }(4,-2)$
  • Foci: $(1\pm\sqrt{5},-2)$
  • Minor-axis endpoints: $(1,-2\pm2)\implies(1,0)\text{ and }(1,-4)$
Exercise • Hyperbola

Exercise: a hyperbola

Watch the negative factor sign, determine the transverse axis, build the guide box, and find the asymptotes.

$$4y^2-9x^2-8y-36x-68=0$$

1. Group & factor — watch the minus sign:

$$4(y^2-2y)-9(x^2+4x)=68$$

Factoring $-9$ from $-36x$ leaves $+4x$.

2. Complete squares and balance the RHS:

$$4(y^2-2y+1)-9(x^2+4x+4)=68+4-36=36$$

Adding $4$ inside $-9(\dots)$ subtracts $36$, so subtract $36$ from RHS.

3. Standard form:

$$\boxed{\frac{(y-1)^2}{9}-\frac{(x+2)^2}{4}=1}$$
Blue: vertices $(-2,4),(-2,-2)$ · orange: foci $(-2,1\pm\sqrt{13})$ · dashed gray: box · dashed teal: asymptotes.
Key features
  • Center: $(-2,1)$ · positive $y\implies$ vertical transverse axis
  • $a=3$ (vert), $b=2$ (horiz) · $c=\sqrt{9+4}=\sqrt{13}\approx3.61$
  • Vertices: $(-2,1\pm3)\implies(-2,4)\text{ and }(-2,-2)$
  • Foci: $(-2,1\pm\sqrt{13})$
  • Asymptotes: slopes $\pm\frac{a}{b}=\pm\frac32\implies\boxed{y-1=\pm\frac32(x+2)}$
§10.5 • Summary

Conic equations and key features

Complete the square → write standard form → find key points → sketch.

ConicStandard form (one orientation)Key points
Parabola$(x-h)^2=4p(y-k)$Focus $(h,k+p)$
Directrix $y=k-p$
Ellipse$\dfrac{(x-h)^2}{a^2}+\dfrac{(y-k)^2}{b^2}=1$
$a>b>0$; horizontal major axis
$c^2=a^2-b^2$
Distance sum $=2a$
Hyperbola$\dfrac{(x-h)^2}{a^2}-\dfrac{(y-k)^2}{b^2}=1$$c^2=a^2+b^2$
$y-k=\pm\dfrac ba(x-h)$
Identify the conic: one squared variable usually gives a parabola; two squared terms with the same sign usually give an ellipse; opposite signs usually give a hyperbola. Check the equation: some cases give lines, a single point, or no real points. These rules assume no $xy$ term.