Conic Sections
Explore circles, ellipses, parabolas, and hyperbolas
You have probably seen circles, ovals, and U-shaped curves many times in your life without thinking much about where they come from. Here is something remarkable: all of these shapes are related. They are all born from the same simple act - slicing a cone with a flat plane. Change the angle of your slice, and you get a different curve. This family of curves, called conic sections, shows up everywhere from the orbits of planets to the shape of satellite dishes.
If you have ever used a flashlight in the dark, you have already seen conic sections in action. Point the flashlight straight at a wall and you get a circle. Tilt it, and the circle stretches into an oval (an ellipse). Tilt it more, and you get a parabola. These are not abstract mathematical inventions - they are shapes that nature and physics use constantly.
Core Concepts
Where Do Conic Sections Come From?
Imagine a double cone - two ice cream cones joined at their tips, extending infinitely in both directions. Now imagine slicing through this double cone with a flat plane.
- Slice horizontally (perpendicular to the axis): You get a circle
- Slice at a moderate angle: You get an ellipse (an oval)
- Slice parallel to the edge of the cone: You get a parabola
- Slice steeply (cutting through both cones): You get a hyperbola (two separate curves)
This is why they are called “conic sections” - they are literally sections (cross-sections) of a cone. The ancient Greeks discovered these curves over 2,000 years ago, and they remain central to mathematics, physics, and engineering today.
Circles: The Perfect Distance
A circle is the simplest conic section. You already know what a circle looks like, but here is the precise definition: a circle is the set of all points that are the same distance from a fixed center point.
That fixed distance is called the radius. The center point is usually written as $(h, k)$.
The standard form of a circle’s equation is:
$$(x - h)^2 + (y - k)^2 = r^2$$
where $(h, k)$ is the center and $r$ is the radius.
This equation comes directly from the distance formula. Any point $(x, y)$ on the circle must be exactly $r$ units from the center $(h, k)$. The distance formula gives us $\sqrt{(x-h)^2 + (y-k)^2} = r$, and squaring both sides gives us the standard form.
Special case: When the center is at the origin $(0, 0)$, the equation simplifies to:
$$x^2 + y^2 = r^2$$
Key features of a circle:
- Center: $(h, k)$
- Radius: $r$
- Diameter: $2r$ (passes through center)
Parabolas: Focus and Directrix
You have seen parabolas before - they are the U-shaped curves from quadratic functions. But there is another way to define a parabola that reveals its deeper nature.
A parabola is the set of all points that are equidistant from a fixed point (called the focus) and a fixed line (called the directrix).
Every point on the parabola is exactly the same distance from the focus as it is from the directrix. This definition might seem strange at first, but it is what gives parabolas their special reflective properties - why satellite dishes and flashlight reflectors are parabolic.
Vertical parabolas (opening up or down):
$$y - k = \frac{1}{4p}(x - h)^2$$
or equivalently:
$$(x - h)^2 = 4p(y - k)$$
where $(h, k)$ is the vertex and $p$ is the directed distance from the vertex to the focus.
- If $p > 0$, the parabola opens upward (focus is above vertex)
- If $p < 0$, the parabola opens downward (focus is below vertex)
Horizontal parabolas (opening left or right):
$$(y - k)^2 = 4p(x - h)$$
- If $p > 0$, the parabola opens right (focus is right of vertex)
- If $p < 0$, the parabola opens left (focus is left of vertex)
Key features of a parabola:
- Vertex: $(h, k)$ - the turning point
- Focus: $p$ units from the vertex in the direction it opens
- Directrix: A line $p$ units from the vertex in the opposite direction
- Axis of symmetry: The line through the vertex and focus
Ellipses: The Stretched Circle
An ellipse looks like a stretched or squashed circle - think of the shape of a running track or a planet’s orbit. But its mathematical definition involves two special points, not one.
An ellipse is the set of all points where the sum of the distances to two fixed points (called foci) is constant.
Imagine hammering two nails into a board, tying a loose string between them, and tracing a curve with a pencil while keeping the string taut. The curve you draw is an ellipse. The nails are the foci.
The standard form of an ellipse centered at $(h, k)$ is:
$$\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1$$
where $a$ and $b$ are the distances from the center to the ellipse along the horizontal and vertical axes.
Horizontal ellipse (wider than tall, $a > b$):
$$\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1$$
- Major axis is horizontal, length $2a$
- Minor axis is vertical, length $2b$
- Vertices: $(h \pm a, k)$
- Co-vertices: $(h, k \pm b)$
- Foci: $(h \pm c, k)$ where $c^2 = a^2 - b^2$
Vertical ellipse (taller than wide, $b > a$):
$$\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1$$
- Major axis is vertical, length $2b$
- Minor axis is horizontal, length $2a$
- Vertices: $(h, k \pm b)$
- Co-vertices: $(h \pm a, k)$
- Foci: $(h, k \pm c)$ where $c^2 = b^2 - a^2$
The relationship $c^2 = |a^2 - b^2|$ is crucial. The value $c$ is the distance from the center to each focus, and it is always less than the larger of $a$ and $b$ (the foci are always inside the ellipse, on the major axis).
Hyperbolas: The Difference of Distances
A hyperbola looks like two separate curved pieces that mirror each other. Unlike an ellipse, which is one connected curve, a hyperbola has two branches.
A hyperbola is the set of all points where the absolute difference of the distances to two fixed points (the foci) is constant.
Notice the key difference from an ellipse: ellipse uses sum, hyperbola uses difference.
Horizontal hyperbola (opens left and right):
$$\frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1$$
- Center: $(h, k)$
- Vertices: $(h \pm a, k)$ - the points closest to the center
- Foci: $(h \pm c, k)$ where $c^2 = a^2 + b^2$
- Transverse axis: The segment connecting the vertices (horizontal, length $2a$)
- Conjugate axis: Perpendicular to transverse axis (vertical, length $2b$)
Vertical hyperbola (opens up and down):
$$\frac{(y - k)^2}{a^2} - \frac{(x - h)^2}{b^2} = 1$$
- Center: $(h, k)$
- Vertices: $(h, k \pm a)$
- Foci: $(h, k \pm c)$ where $c^2 = a^2 + b^2$
Asymptotes: Hyperbolas have two diagonal lines that the branches approach but never touch. For a hyperbola centered at the origin:
- Horizontal hyperbola: $y = \pm\frac{b}{a}x$
- Vertical hyperbola: $y = \pm\frac{a}{b}x$
For hyperbolas centered at $(h, k)$, the asymptotes pass through the center with the same slopes.
How to Tell Which Conic You Have
When given an equation, you can identify the conic by looking at its structure:
| Conic | Distinguishing Feature |
|---|---|
| Circle | $x^2$ and $y^2$ have the same coefficient |
| Ellipse | $x^2$ and $y^2$ have different positive coefficients |
| Parabola | Only one variable is squared |
| Hyperbola | $x^2$ and $y^2$ have opposite signs |
Eccentricity: Measuring “Ovalness”
Eccentricity (denoted $e$) is a number that describes how “stretched” a conic section is.
- Circle: $e = 0$ (not stretched at all)
- Ellipse: $0 < e < 1$ (the closer to 0, the more circular; the closer to 1, the more elongated)
- Parabola: $e = 1$
- Hyperbola: $e > 1$ (the larger, the more “open” the branches)
For ellipses and hyperbolas, eccentricity is calculated as:
$$e = \frac{c}{a}$$
where $c$ is the distance from center to focus and $a$ is the distance from center to vertex.
Earth’s orbit around the sun has eccentricity $e \approx 0.017$, which is very close to circular. Halley’s Comet has eccentricity $e \approx 0.967$, making its orbit a highly elongated ellipse.
Notation and Terminology
| Term | Meaning | Example |
|---|---|---|
| Circle | All points equidistant from center | $(x-2)^2 + (y+1)^2 = 16$ |
| Ellipse | Sum of distances to foci is constant | $\frac{x^2}{25} + \frac{y^2}{9} = 1$ |
| Hyperbola | Difference of distances to foci is constant | $\frac{x^2}{16} - \frac{y^2}{9} = 1$ |
| Focus (foci) | Special point(s) defining the curve | For ellipse: $(\pm c, 0)$ |
| Directrix | Line used with focus to define parabola | $y = -p$ for upward parabola |
| Vertices | Key points on the conic | Endpoints of major axis |
| Asymptotes | Lines hyperbola approaches | $y = \pm\frac{b}{a}x$ |
| Major axis | Longest diameter of ellipse | Length $2a$ |
| Minor axis | Shortest diameter of ellipse | Length $2b$ |
| Transverse axis | Segment connecting hyperbola’s vertices | Length $2a$ |
| Eccentricity | Measure of how “stretched” the conic is | $e = \frac{c}{a}$ |
Examples
Write the equation of a circle with center $(3, -2)$ and radius 5.
Solution:
The standard form of a circle is: $$(x - h)^2 + (y - k)^2 = r^2$$
We have:
- Center $(h, k) = (3, -2)$
- Radius $r = 5$
Substitute into the formula: $$(x - 3)^2 + (y - (-2))^2 = 5^2$$
Simplify: $$(x - 3)^2 + (y + 2)^2 = 25$$
Check: The center is at $(3, -2)$. Let us verify the point $(3, 3)$ is on the circle (it should be 5 units above the center): $$(3 - 3)^2 + (3 + 2)^2 = 0 + 25 = 25$$ ✓
Key insight: Notice how $(y - (-2))$ becomes $(y + 2)$. The signs in the equation are always opposite to the coordinates of the center.
Identify the center and radius of the circle $(x + 1)^2 + (y - 4)^2 = 36$.
Solution:
Compare to the standard form $(x - h)^2 + (y - k)^2 = r^2$.
From $(x + 1)^2$: This is $(x - (-1))^2$, so $h = -1$
From $(y - 4)^2$: This is $(y - 4)^2$, so $k = 4$
From $= 36$: This is $r^2 = 36$, so $r = 6$
The center is $(-1, 4)$ and the radius is $6$.
Common mistake: Students often get the signs wrong. Remember: $(x + 1)^2$ means $h = -1$, not $h = 1$. The sign in the equation is opposite to the coordinate.
Visualization: This circle is centered 1 unit left of the y-axis and 4 units above the x-axis, with points extending 6 units in every direction from that center.
Graph the ellipse $\frac{x^2}{16} + \frac{y^2}{25} = 1$ and find its foci.
Solution:
Step 1: Identify the key values.
The equation is already in standard form: $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$
Comparing: $a^2 = 16$ and $b^2 = 25$, so $a = 4$ and $b = 5$
Since $b > a$ (25 > 16), this is a vertical ellipse (taller than wide).
Step 2: Find the center.
The equation has no $(x - h)$ or $(y - k)$ shifts, so the center is at the origin $(0, 0)$.
Step 3: Find the vertices and co-vertices.
For a vertical ellipse:
- Vertices (on major axis): $(0, \pm b) = (0, \pm 5)$, so $(0, 5)$ and $(0, -5)$
- Co-vertices (on minor axis): $(\pm a, 0) = (\pm 4, 0)$, so $(4, 0)$ and $(-4, 0)$
Step 4: Find the foci.
For a vertical ellipse, $c^2 = b^2 - a^2$: $$c^2 = 25 - 16 = 9$$ $$c = 3$$
The foci are at $(0, \pm c) = (0, \pm 3)$, so $(0, 3)$ and $(0, -3)$.
Step 5: Sketch the graph.
Plot the center, vertices, co-vertices, and foci. Draw a smooth oval passing through the vertices and co-vertices.
Verification: The sum of distances from any point on the ellipse to both foci should equal $2b = 10$. From the point $(4, 0)$:
- Distance to $(0, 3)$: $\sqrt{16 + 9} = 5$
- Distance to $(0, -3)$: $\sqrt{16 + 9} = 5$
- Sum: $5 + 5 = 10$ ✓
Write the equation of a parabola with vertex at the origin and focus at $(0, 3)$.
Solution:
Step 1: Determine the orientation.
The vertex is at $(0, 0)$ and the focus is at $(0, 3)$.
Since the focus is directly above the vertex (same x-coordinate, larger y-coordinate), this parabola opens upward.
Step 2: Find $p$.
The value $p$ is the directed distance from vertex to focus. $$p = 3 - 0 = 3$$
Since the parabola opens upward, $p > 0$, which checks out.
Step 3: Write the equation.
For a vertical parabola with vertex at origin, the standard form is: $$x^2 = 4py$$
Substitute $p = 3$: $$x^2 = 4(3)y = 12y$$
The equation is $x^2 = 12y$ or equivalently $y = \frac{x^2}{12}$.
Step 4: Find the directrix.
The directrix is $p$ units below the vertex (opposite direction from focus): $$y = -p = -3$$
Check: Take a point on the parabola, say where $x = 6$: $$y = \frac{36}{12} = 3$$
The point $(6, 3)$ should be equidistant from focus $(0, 3)$ and directrix $y = -3$:
- Distance to focus: $\sqrt{(6-0)^2 + (3-3)^2} = 6$
- Distance to directrix: $3 - (-3) = 6$ ✓
Graph the hyperbola $\frac{(x-1)^2}{9} - \frac{(y+2)^2}{16} = 1$ and find its vertices, foci, and asymptotes.
Solution:
Step 1: Identify the form and orientation.
The equation has the form $\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1$.
Since the positive term contains $x$, this is a horizontal hyperbola (opens left and right).
Step 2: Identify the key values.
- Center: $(h, k) = (1, -2)$
- $a^2 = 9$, so $a = 3$
- $b^2 = 16$, so $b = 4$
Step 3: Find the vertices.
For a horizontal hyperbola, vertices are $a$ units left and right of center: $$\text{Vertices: } (h \pm a, k) = (1 \pm 3, -2)$$
So the vertices are $(4, -2)$ and $(-2, -2)$.
Step 4: Find the foci.
For hyperbolas, $c^2 = a^2 + b^2$ (note: addition, not subtraction like ellipses): $$c^2 = 9 + 16 = 25$$ $$c = 5$$
The foci are at $(h \pm c, k) = (1 \pm 5, -2)$:
So the foci are $(6, -2)$ and $(-4, -2)$.
Step 5: Find the asymptotes.
For a horizontal hyperbola centered at $(h, k)$, the asymptotes have slopes $\pm\frac{b}{a}$ and pass through the center:
$$y - k = \pm\frac{b}{a}(x - h)$$ $$y - (-2) = \pm\frac{4}{3}(x - 1)$$ $$y + 2 = \pm\frac{4}{3}(x - 1)$$
The two asymptotes are:
- $y = \frac{4}{3}(x - 1) - 2 = \frac{4}{3}x - \frac{4}{3} - 2 = \frac{4}{3}x - \frac{10}{3}$
- $y = -\frac{4}{3}(x - 1) - 2 = -\frac{4}{3}x + \frac{4}{3} - 2 = -\frac{4}{3}x - \frac{2}{3}$
Step 6: Sketch the graph.
- Plot the center $(1, -2)$
- Plot the vertices $(4, -2)$ and $(-2, -2)$
- Draw the asymptotes as dashed lines through the center
- Draw the two branches of the hyperbola, each passing through a vertex and approaching (but never touching) the asymptotes
Eccentricity: $e = \frac{c}{a} = \frac{5}{3} \approx 1.67$
Identify the conic and rewrite in standard form: $4x^2 + 9y^2 - 16x + 18y - 11 = 0$
Solution:
Step 1: Identify the conic type.
Both $x^2$ and $y^2$ are present with positive coefficients (4 and 9), but the coefficients are different.
This means it is an ellipse.
Step 2: Group and prepare to complete the square.
Group the $x$ terms and $y$ terms: $$(4x^2 - 16x) + (9y^2 + 18y) = 11$$
Factor out the leading coefficients from each group: $$4(x^2 - 4x) + 9(y^2 + 2y) = 11$$
Step 3: Complete the square for each variable.
For $x^2 - 4x$: Take half of $-4$, which is $-2$. Square it: $(-2)^2 = 4$
For $y^2 + 2y$: Take half of $2$, which is $1$. Square it: $(1)^2 = 1$
Add these inside the parentheses, but remember to balance the equation: $$4(x^2 - 4x + 4) + 9(y^2 + 2y + 1) = 11 + 4(4) + 9(1)$$
We added $4 \times 4 = 16$ and $9 \times 1 = 9$ to the left side, so add those to the right: $$4(x - 2)^2 + 9(y + 1)^2 = 11 + 16 + 9 = 36$$
Step 4: Divide to get standard form.
Divide everything by 36: $$\frac{4(x - 2)^2}{36} + \frac{9(y + 1)^2}{36} = 1$$
Simplify: $$\frac{(x - 2)^2}{9} + \frac{(y + 1)^2}{4} = 1$$
Step 5: Interpret the result.
This is an ellipse with:
- Center: $(2, -1)$
- $a^2 = 9$, so $a = 3$ (horizontal direction)
- $b^2 = 4$, so $b = 2$ (vertical direction)
Since $a > b$, this is a horizontal ellipse (wider than tall).
- Vertices: $(2 \pm 3, -1) = (5, -1)$ and $(-1, -1)$
- Co-vertices: $(2, -1 \pm 2) = (2, 1)$ and $(2, -3)$
- $c^2 = a^2 - b^2 = 9 - 4 = 5$, so $c = \sqrt{5}$
- Foci: $(2 \pm \sqrt{5}, -1)$
Check: Substitute the center $(2, -1)$ back into the original equation - wait, the center is not on the ellipse. Let us check a vertex, $(5, -1)$: $$4(5)^2 + 9(-1)^2 - 16(5) + 18(-1) - 11$$ $$= 100 + 9 - 80 - 18 - 11 = 0$$ ✓
Key Properties and Rules
Circle Properties
For $(x - h)^2 + (y - k)^2 = r^2$:
- All points are distance $r$ from center $(h, k)$
- Circumference: $2\pi r$
- Area: $\pi r^2$
Parabola Properties
For $(x - h)^2 = 4p(y - k)$ (vertical):
- Vertex: $(h, k)$
- Focus: $(h, k + p)$
- Directrix: $y = k - p$
- Axis of symmetry: $x = h$
- If $p > 0$, opens up; if $p < 0$, opens down
For $(y - k)^2 = 4p(x - h)$ (horizontal):
- Vertex: $(h, k)$
- Focus: $(h + p, k)$
- Directrix: $x = h - p$
- Axis of symmetry: $y = k$
- If $p > 0$, opens right; if $p < 0$, opens left
Ellipse Properties
For $\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$:
- Center: $(h, k)$
- If $a > b$: horizontal ellipse, $c^2 = a^2 - b^2$, foci at $(h \pm c, k)$
- If $b > a$: vertical ellipse, $c^2 = b^2 - a^2$, foci at $(h, k \pm c)$
- Sum of distances from any point to both foci $= 2 \times (\text{larger of } a, b)$
Hyperbola Properties
For $\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1$ (horizontal):
- Center: $(h, k)$
- Vertices: $(h \pm a, k)$
- Foci: $(h \pm c, k)$ where $c^2 = a^2 + b^2$
- Asymptotes: $y - k = \pm\frac{b}{a}(x - h)$
For $\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1$ (vertical):
- Center: $(h, k)$
- Vertices: $(h, k \pm a)$
- Foci: $(h, k \pm c)$ where $c^2 = a^2 + b^2$
- Asymptotes: $y - k = \pm\frac{a}{b}(x - h)$
Identifying Conics from General Form
The general second-degree equation is $Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$.
When $B = 0$ (no $xy$ term):
- Circle: $A = C$ (same coefficient on both squared terms)
- Ellipse: $A$ and $C$ have the same sign but $A \neq C$
- Hyperbola: $A$ and $C$ have opposite signs
- Parabola: Either $A = 0$ or $C = 0$ (only one variable squared)
Common Mistakes to Avoid
- Confusing the relationships: Ellipse uses $c^2 = a^2 - b^2$, but hyperbola uses $c^2 = a^2 + b^2$
- Signs in standard form: $(x + 3)^2$ means the center has $h = -3$, not $+3$
- Which is $a$ and which is $b$: For ellipses, $a$ is always the larger value; for hyperbolas, $a^2$ is always under the positive term
- Forgetting to complete the square properly: When you factor out a coefficient, remember to multiply what you add inside by that coefficient on the other side
Real-World Applications
Planetary Orbits (Ellipses)
Johannes Kepler discovered in the 1600s that planets orbit the sun in ellipses, not perfect circles. The sun sits at one focus of each planet’s elliptical orbit. Earth’s orbit is nearly circular (eccentricity 0.017), but comets often have highly elongated elliptical orbits.
Satellite Dishes and Flashlights (Parabolas)
The parabola has a remarkable reflective property: any signal coming parallel to the axis of symmetry reflects off the parabola and passes through the focus. Satellite dishes are parabolic because they concentrate incoming radio waves at a single point (the receiver at the focus). Flashlights work the opposite way - a bulb at the focus sends light outward in parallel rays.
Whispering Galleries (Ellipses)
In an elliptical room, a whisper at one focus can be heard clearly at the other focus, even across a large distance. This happens because sound waves bouncing off the elliptical walls all converge at the other focus. The National Statuary Hall in the U.S. Capitol building demonstrates this effect.
Cooling Towers (Hyperbolas)
The distinctive hourglass shape of nuclear power plant cooling towers is a hyperbola rotated around its axis. This shape, called a hyperboloid, is structurally strong and efficient for air circulation.
GPS Navigation (Hyperbolic Positioning)
GPS uses the difference in arrival times of signals from multiple satellites. If you know the difference in distances to two satellites, you are somewhere on a hyperbola with those satellites as foci. The intersection of multiple hyperbolas pinpoints your location.
Self-Test Problems
Problem 1: Write the equation of a circle with center $(-2, 5)$ and radius 7.
Show Answer
Using the standard form $(x - h)^2 + (y - k)^2 = r^2$ with center $(-2, 5)$ and radius 7:
$$(x - (-2))^2 + (y - 5)^2 = 7^2$$ $$(x + 2)^2 + (y - 5)^2 = 49$$
Problem 2: What type of conic is $3x^2 + 3y^2 - 6x + 12y - 15 = 0$?
Show Answer
Both $x^2$ and $y^2$ have the same coefficient (3), so this is a circle.
To find its center and radius, divide by 3 and complete the square: $$x^2 + y^2 - 2x + 4y - 5 = 0$$ $$(x^2 - 2x + 1) + (y^2 + 4y + 4) = 5 + 1 + 4$$ $$(x - 1)^2 + (y + 2)^2 = 10$$
Center: $(1, -2)$, Radius: $\sqrt{10}$
Problem 3: Find the foci of the ellipse $\frac{x^2}{36} + \frac{y^2}{20} = 1$.
Show Answer
Here $a^2 = 36$ and $b^2 = 20$, so $a = 6$ and $b = \sqrt{20} = 2\sqrt{5}$.
Since $a > b$, this is a horizontal ellipse.
Find $c$: $c^2 = a^2 - b^2 = 36 - 20 = 16$, so $c = 4$
The foci are at $(\pm c, 0) = (\pm 4, 0)$, which means $(4, 0)$ and $(-4, 0)$.
Problem 4: Write the equation of a parabola with vertex $(2, -1)$ and focus $(2, 2)$.
Show Answer
The focus is directly above the vertex (same x-coordinate), so this is a vertical parabola opening upward.
The distance from vertex to focus is $p = 2 - (-1) = 3$.
Using the form $(x - h)^2 = 4p(y - k)$: $$(x - 2)^2 = 4(3)(y - (-1))$$ $$(x - 2)^2 = 12(y + 1)$$
Problem 5: Find the asymptotes of the hyperbola $\frac{y^2}{9} - \frac{x^2}{4} = 1$.
Show Answer
This is a vertical hyperbola (the positive term has $y^2$) centered at the origin.
Here $a^2 = 9$ (under $y^2$) and $b^2 = 4$ (under $x^2$), so $a = 3$ and $b = 2$.
For a vertical hyperbola centered at origin, the asymptotes are $y = \pm\frac{a}{b}x$:
$$y = \pm\frac{3}{2}x$$
The asymptotes are $y = \frac{3}{2}x$ and $y = -\frac{3}{2}x$.
Problem 6: Convert $x^2 - 4y^2 + 6x + 8y + 1 = 0$ to standard form and identify the conic.
Show Answer
The coefficients of $x^2$ and $y^2$ have opposite signs (1 and -4), so this is a hyperbola.
Group and complete the square: $$(x^2 + 6x) - 4(y^2 - 2y) = -1$$
Complete the square for each:
- $x^2 + 6x$: add $(6/2)^2 = 9$
- $y^2 - 2y$: add $(-2/2)^2 = 1$
$$(x^2 + 6x + 9) - 4(y^2 - 2y + 1) = -1 + 9 - 4(1)$$ $$(x + 3)^2 - 4(y - 1)^2 = 4$$
Divide by 4: $$\frac{(x + 3)^2}{4} - \frac{(y - 1)^2}{1} = 1$$
This is a horizontal hyperbola with center $(-3, 1)$, $a = 2$, and $b = 1$.
Problem 7: An ellipse has foci at $(\pm 4, 0)$ and vertices at $(\pm 5, 0)$. Write its equation.
Show Answer
From the vertices, $a = 5$. From the foci, $c = 4$.
For an ellipse, $c^2 = a^2 - b^2$: $$16 = 25 - b^2$$ $$b^2 = 9$$
The ellipse is centered at the origin with horizontal major axis: $$\frac{x^2}{25} + \frac{y^2}{9} = 1$$
Summary
- Conic sections are curves formed by slicing a cone with a plane: circles, ellipses, parabolas, and hyperbolas.
- A circle is all points equidistant from a center. Standard form: $(x - h)^2 + (y - k)^2 = r^2$.
- A parabola is all points equidistant from a focus and directrix. The equation involves only one squared variable.
- An ellipse is all points where the sum of distances to two foci is constant. Standard form: $\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$. Find foci using $c^2 = |a^2 - b^2|$.
- A hyperbola is all points where the difference of distances to two foci is constant. It has two branches and asymptotes. Find foci using $c^2 = a^2 + b^2$.
- To identify a conic from general form: same coefficients means circle, same sign but different coefficients means ellipse, opposite signs means hyperbola, only one squared variable means parabola.
- Eccentricity measures how stretched a conic is: $e = 0$ for circles, $0 < e < 1$ for ellipses, $e = 1$ for parabolas, $e > 1$ for hyperbolas.
- Conic sections appear everywhere in the real world: planetary orbits, satellite dishes, whispering galleries, cooling towers, and GPS navigation all rely on these curves.
You now have the tools to recognize, graph, and write equations for all four types of conic sections. These curves are not just abstract mathematical objects - they are the shapes that govern how planets move, how sound travels, and how engineers design structures. When you see a parabolic dish or learn about an elliptical orbit, you will understand the mathematics behind it.