Conic Sections

Study Sheet

Conic Sections

Circles, parabolas, ellipses & hyperbolas — with worked examples and labeled graphs

The Four Conics & Review Formulas

Concept
Big idea

A conic section is a curve you get by slicing a double cone with a flat plane. Tilting the plane a little more each time produces, in order, a circle, an ellipse, a parabola, and a hyperbola. Every one of them can be written as a single second-degree equation

Ax2+Cy2+Dx+Ey+F=0,Ax^2+Cy^2+Dx+Ey+F=0,

and the whole topic is really one question asked four ways: what shape does this equation draw, and where are its key points?

Slicing a cone gives four curves — circle, ellipse, parabola, hyperbola — and each has a defining distance property: an ellipse is the set of points whose distances to two foci sum to a constant, a hyperbola where they differ by a constant, a parabola where the distance to a focus equals the distance to a line. Completing the square turns any second-degree equation into a standard form that reveals which curve it is.

Every conic is defined by distances, which is why the distance formula is the key tool: a circle is all points at one distance from a center; an ellipse, all points whose distances to two foci add to a constant; a hyperbola, all points whose distances differ by a constant; a parabola, all points equally far from a focus and a line. Writing any of these with the distance formula and squaring produces a second-degree equation.

Concept
Two tools you already own (review)

For points (x1,y1)(x_1,y_1) and (x2,y2)(x_2,y_2):

Distance: d=(x2−x1)2+(y2−y1)2,Midpoint: M=(x1+x22, y1+y22).\textbf{Distance: } d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2},\qquad \textbf{Midpoint: } M=\left(\dfrac{x_1+x_2}{2},\ \dfrac{y_1+y_2}{2}\right).

The distance formula is the circle equation in disguise (a circle is all points a fixed distance from the center), so keep it close.

Example
Example: distance and midpoint
xy(−3, −2)(5, 4)M(1, 1)86
Legs 88 and 66 give distance 1010; the midpoint averages the endpoints.

For (−3,−2)(-3,-2) and (5,4)(5,4):

d=(5−(−3))2+(4−(−2))2=82+62=100=10,d=\sqrt{(5-(-3))^2+(4-(-2))^2}=\sqrt{8^2+6^2}=\sqrt{100}=10,
M=(−3+52,−2+42)=(1,1).M=\left(\tfrac{-3+5}{2},\tfrac{-2+4}{2}\right)=(1,1).
Side note
Where it goes wrong

Reading x29+y24=1\tfrac{x^2}{9} + \tfrac{y^2}{4} = 1 as an ellipse with semi-axes 99 and 44. The semi-axes are 9=3\sqrt9 = 3 and 4=2\sqrt4 = 2.

The four conics, side by side.

Tip

Tip: A circle is just a special ellipse where both radii are equal. If you understand the ellipse deeply, the circle is free.

Try it
Try it: distance and midpoint

For A(−4,7)A(-4, 7) and B(2,−1)B(2, -1), find ABAB and the midpoint. Is (2,7)(2, 7) on the circle with diameter ABAB?

Show the answer

AB=62+82=10AB = \sqrt{6^2 + 8^2} = 10. Midpoint (−1,3)(-1, 3), so the circle has center (−1,3)(-1, 3) and radius 55. Distance from (−1,3)(-1, 3) to (2,7)(2, 7) is 9+16=5\sqrt{9 + 16} = 5: yes, it is on the circle.

Circles

Concept
Standard equation of a circle

A circle with center (h,k)(h,k) and radius rr is the set of all points a distance rr from the center:

(x−h)2+(y−k)2=r2.(x-h)^2+(y-k)^2=r^2.

Read the center straight off the equation as the opposites of the numbers inside the parentheses, and take r= r2 r=\sqrt{\,r^2\,}.

The circle equation is the distance formula with the distance fixed: a point (x,y)(x, y) is on the circle exactly when its distance to (h,k)(h, k) is rr, that is (x−h)2+(y−k)2=r\sqrt{(x - h)^2 + (y - k)^2} = r; squaring gives the standard form. So the center shows up with flipped signs for the same reason a vertex does in (x−h)2(x - h)^2.

Example
Example: center and radius
xy(3, −2)5
Center (3,−2)(3, -2), radius 55: the point (7,1)(7, 1) is 55 away.

For (x−3)2+(y+2)2=25(x-3)^2+(y+2)^2=25 the center is (3,−2)(3,-2) (note y+2=y−(−2)y+2=y-(-2)) and the radius is r=25=5r=\sqrt{25}=5.

Side note
Where it goes wrong

Reading the radius as r2r^2: (x−3)2+(y+2)2=25(x - 3)^2 + (y + 2)^2 = 25 has radius 25=5\sqrt{25} = 5, not 2525.

Example
Example: completing the square (general form →\to standard form)

Convert x2+y2−6x+4y−12=0x^2+y^2-6x+4y-12=0 into standard form.

Group the xx's and yy's and move the constant:

(x2−6x)+(y2+4y)=12.(x^2-6x)+(y^2+4y)=12.

Complete each square by adding (−62)2=9\left(\tfrac{-6}{2}\right)^2=9 and (42)2=4\left(\tfrac{4}{2}\right)^2=4 to both sides:

(x2−6x+9)+(y2+4y+4)=12+9+4,(x^2-6x+9)+(y^2+4y+4)=12+9+4,
(x−3)2+(y+2)2=25.(x-3)^2+(y+2)^2=25.

So the center is (3,−2)(3,-2) and the radius is r=5r=5.

Tip

Tip: When you complete the square, whatever you add on the left you must add on the right. A common slip is adding 99 and 44 only on one side.

Try it
Try it: to standard form

Find the center and radius of x2+y2+8x−2y+8=0x^2 + y^2 + 8x - 2y + 8 = 0.

Show the answer

Group and complete squares: (x2+8x+16)+(y2−2y+1)=−8+16+1(x^2 + 8x + 16) + (y^2 - 2y + 1) = -8 + 16 + 1, so (x+4)2+(y−1)2=9(x + 4)^2 + (y - 1)^2 = 9. Center (−4,1)(-4, 1), radius 33.

Parabolas as Conics

Concept
Vertex form and direction of opening

A parabola is every point equidistant from a fixed focus and a fixed line called the directrix. In vertex form:

y=a(x−h)2+kopens up (a>0) or down (a<0),y=a(x-h)^2+k \quad\text{opens up } (a>0)\text{ or down } (a<0),
x=a(y−k)2+hopens right (a>0) or left (a<0).x=a(y-k)^2+h \quad\text{opens right } (a>0)\text{ or left } (a<0).

Either way the vertex is (h,k)(h,k): it is the turning point, halfway between the focus and the directrix.

Take the focus (0,p)(0, p) and the directrix y=−py = -p. A point (x,y)(x, y) is equally far from both when x2+(y−p)2=y+p\sqrt{x^2 + (y - p)^2} = y + p; squaring and simplifying leaves x2=4pyx^2 = 4py, or y=14px2y = \tfrac{1}{4p}x^2. That is why the coefficient reveals pp: for y=14x2y = \tfrac14 x^2, 4p=44p = 4 and the focus is 11 unit above the vertex.

Concept
Focus and directrix (intro)

Write an upward parabola as y=14p(x−h)2+ky=\dfrac{1}{4p}(x-h)^2+k. Then pp is the distance from the vertex to the focus (and to the directrix):

focus (h, k+p),directrix y=k−p.\textbf{focus }(h,\,k+p),\qquad \textbf{directrix } y=k-p.

The focus sits inside the curve; the directrix is the same distance out the other side.

Example
Example: vertex, focus, directrix
xyfocus (0, 1)22y = −1
The point (2,1)(2, 1) is 22 from the focus and 22 from the directrix y=−1y = -1.

For y=14x2y=\dfrac{1}{4}x^2 we have 14p=14\dfrac{1}{4p}=\dfrac14, so 4p=44p=4 and p=1p=1. The vertex is (0,0)(0,0), the parabola opens up, the focus is (0,1)(0,1), and the directrix is the line y=−1y=-1.

Side note
Where it goes wrong

Reading pp as the coefficient: for y=14x2y = \tfrac14 x^2, "focus at (0,14)(0, \tfrac14)." The coefficient is 14p\tfrac{1}{4p}; setting it equal to 14\tfrac14 gives p=1p = 1, so the focus is (0,1)(0, 1).

Tip

Tip: An x2x^2 term (and no y2y^2) means the parabola opens up or down; a y2y^2 term (and no x2x^2) means it opens left or right. A parabola is the only conic with just one squared variable.

Try it
Try it: focus and directrix

Find the vertex, focus and directrix of y=18(x−2)2+3y = \tfrac18(x - 2)^2 + 3.

Show the answer

Vertex (2,3)(2, 3). 14p=18\tfrac{1}{4p} = \tfrac18 gives p=2p = 2, and it opens up, so the focus is (2,5)(2, 5) and the directrix is y=1y = 1.

Ellipses

Concept
Standard equation of an ellipse (center at origin)
x2a2+y2b2=1,a>b>0.\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1,\qquad a>b>0.

The larger denominator sits under the axis holding the major axis (the long one). Let aa be the larger of the two, bb the smaller. Then:

  • Vertices: endpoints of the major axis, distance aa from center.
  • Co-vertices: endpoints of the minor axis, distance bb from center.
  • Foci: distance cc from center along the major axis, where c2=a2−b2c^2=a^2-b^2.

A shifted ellipse uses (x−h)2a2+(y−k)2b2=1\dfrac{(x-h)^2}{a^2}+\dfrac{(y-k)^2}{b^2}=1 with center (h,k)(h,k).

An ellipse is all points whose distances to the two foci add to 2a2a, the length of the major axis. At the top of the ellipse the two distances are equal, each aa, and they form a right triangle with legs bb and cc; so a2=b2+c2a^2 = b^2 + c^2, which is where c2=a2−b2c^2 = a^2 - b^2 comes from. For x225+y29=1\tfrac{x^2}{25} + \tfrac{y^2}{9} = 1: 5=32+425 = \sqrt{3^2 + 4^2}.

Example
Example: full feature analysis
xyF(4, 0)F(−4, 0)55
From the top vertex (0,3)(0, 3) each focus is 55 away: 5+5=10=2a5 + 5 = 10 = 2a.

Analyze x225+y29=1\dfrac{x^2}{25}+\dfrac{y^2}{9}=1.

The larger denominator (2525) is under x2x^2, so the major axis is horizontal with a=25=5a=\sqrt{25}=5 and b=9=3b=\sqrt{9}=3. Then c2=a2−b2=25−9=16c^2=a^2-b^2=25-9=16, so c=4c=4.

  • Center (0,0)(0,0);   Vertices (±5,0)(\pm5,0);   Co-vertices (0,±3)(0,\pm3);   Foci (±4,0)(\pm4,0).
Side note
Where it goes wrong

Using the hyperbola relation: c2=a2+b2=34c^2 = a^2 + b^2 = 34, putting the foci outside the ellipse at ±34≈±5.8\pm\sqrt{34} \approx \pm 5.8. For an ellipse the foci are inside: c2=a2−b2=16c^2 = a^2 - b^2 = 16.

Tip

Tip: For an ellipse a>ba>b and the foci hug the long axis, so c<ac<a always. If you ever get c2<0c^2<0, you mislabeled aa and bb.

Try it
Try it: features

For x216+y236=1\dfrac{x^2}{16} + \dfrac{y^2}{36} = 1, find the orientation, vertices, co-vertices and foci.

Show the answer

The larger denominator, 3636, is under y2y^2: the major axis is vertical, a=6a = 6, b=4b = 4. Vertices (0,±6)(0, \pm 6), co-vertices (±4,0)(\pm 4, 0). c2=36−16=20c^2 = 36 - 16 = 20, so foci (0,±25)(0, \pm 2\sqrt5).

Hyperbolas

Concept
Standard equation of a hyperbola (center at origin)
x2a2−y2b2=1  (opens left–right),y2a2−x2b2=1  (opens up–down).\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1 \ \ (\text{opens left--right}),\qquad \dfrac{y^2}{a^2}-\dfrac{x^2}{b^2}=1 \ \ (\text{opens up--down}).

The positive term tells you which way it opens; aa is always the number under the positive term.

  • Vertices: distance aa from center, on the opening axis.
  • Foci: distance cc from center, where c2=a2+b2c^2=a^2+b^2 (note the ++!).
  • Asymptotes: for x2a2−y2b2=1\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1 they are y=±baxy=\pm\dfrac{b}{a}x; for the up–down form they are y=±abxy=\pm\dfrac{a}{b}x.

Far from the center, the 11 in x29−y216=1\tfrac{x^2}{9} - \tfrac{y^2}{16} = 1 becomes negligible next to the squared terms, so the curve nearly satisfies x29=y216\tfrac{x^2}{9} = \tfrac{y^2}{16}, the lines y=±43xy = \pm\tfrac43 x. Those are the asymptotes, the diagonals of the 66-by-88 box. The foci lie beyond the vertices, and c2=a2+b2c^2 = a^2 + b^2 is the box's half-diagonal: 5=9+165 = \sqrt{9 + 16}.

Example
Example: full feature analysis
xyF(5, 0)F(−5, 0)
The asymptotes are the diagonals of the 6×86 \times 8 box; the foci (±5,0)(\pm5, 0) lie at the box's half-diagonal.

Analyze x29−y216=1\dfrac{x^2}{9}-\dfrac{y^2}{16}=1.

The x2x^2 term is positive, so it opens left–right with a=9=3a=\sqrt{9}=3 and b=16=4b=\sqrt{16}=4. Then c2=a2+b2=9+16=25c^2=a^2+b^2=9+16=25, so c=5c=5.

  • Center (0,0)(0,0);   Vertices (±3,0)(\pm3,0);   Foci (±5,0)(\pm5,0);   Asymptotes y=±43xy=\pm\dfrac{4}{3}x.
Side note
Where it goes wrong

Taking the asymptote slopes upside down: y=±34xy = \pm\tfrac34 x for x29−y216=1\tfrac{x^2}{9} - \tfrac{y^2}{16} = 1. The slope is rise over run of the box, ba=43\tfrac ba = \tfrac43.

Tip

Tip: Ellipse uses c2=a2−b2c^2=a^2-b^2; hyperbola uses c2=a2+b2c^2=a^2+b^2. For a hyperbola the foci are farther out than the vertices, so c>ac>a.

Try it
Try it: features

For y24−x221=1\dfrac{y^2}{4} - \dfrac{x^2}{21} = 1, find the direction it opens, the vertices, the foci and the asymptotes.

Show the answer

The y2y^2 term is positive, so it opens up and down; a=2a = 2, b=21b = \sqrt{21}. Vertices (0,±2)(0, \pm 2). c2=4+21=25c^2 = 4 + 21 = 25, foci (0,±5)(0, \pm 5). Asymptotes y=±221xy = \pm\tfrac{2}{\sqrt{21}}x.

Identifying a Conic from General Form

Concept
The classification test

Given Ax2+Cy2+Dx+Ey+F=0Ax^2+Cy^2+Dx+Ey+F=0 (no xyxy term), look only at AA and CC:

  • Circle: A=CA=C (same coefficient on x2x^2 and y2y^2).
  • Parabola: exactly one of A,CA,C is 00 (only one squared term).
  • Ellipse: A≠CA\ne C but AA and CC have the same sign.
  • Hyperbola: AA and CC have opposite signs.

Completing the square never changes the x2x^2 and y2y^2 coefficients, so they already decide the shape: equal coefficients give a circle, same signs give an ellipse, opposite signs a hyperbola, and a missing square a parabola. The linear terms only move the center.

Example
Example: classify each
  • x2+y2−4x=1x^2+y^2-4x=1: A=C=1⇒A=C=1 \Rightarrow circle.
  • y=3x2−xy=3x^2-x (only x2x^2): parabola.
  • 4x2+9y2=364x^2+9y^2=36: A=4, C=9A=4,\,C=9 same sign, unequal ⇒\Rightarrow ellipse.
  • 9x2−4y2=369x^2-4y^2=36: A=9, C=−4A=9,\,C=-4 opposite signs ⇒\Rightarrow hyperbola.
Side note
Where it goes wrong

Calling 4x2+4y2−8x=04x^2 + 4y^2 - 8x = 0 an ellipse because its coefficients are not 11. Equal x2x^2 and y2y^2 coefficients make a circle: dividing by 44 gives (x−1)2+y2=1(x - 1)^2 + y^2 = 1.

Tip

Tip (how to tell them apart at a glance): One squared term ⇒\Rightarrow parabola. Two squared terms: same number ⇒\Rightarrow circle, added ⇒\Rightarrow ellipse, subtracted ⇒\Rightarrow hyperbola.

Try it
Try it: classify

Classify 2x2−3y2+4x=52x^2 - 3y^2 + 4x = 5, x2+6x−y=0x^2 + 6x - y = 0 and 5x2+5y2=205x^2 + 5y^2 = 20.

Show the answer

Opposite signs on the squares: hyperbola. Only x2x^2: parabola. Equal coefficients: circle (radius 22).

Applications (light)

Concept
Where conics show up
  • Ellipses — orbits: planets and satellites travel in ellipses with the larger body at one focus (Kepler's First Law).
  • Ellipses — whispering galleries: a sound starting at one focus reflects off the elliptical ceiling and gathers at the other focus, so a whisper carries across the room.
  • Parabolas — dishes and headlights: a satellite dish or reflector collects incoming rays at its focus (and a bulb at the focus sends out a straight beam).
  • Hyperbolas — navigation and comets: some comets swing past the sun on an open hyperbolic path and never return.

Each application uses a conic's focus. A parabolic dish sends every incoming parallel ray to its focus because each point is as far from the focus as from the directrix, which makes the reflection angles equal. An ellipse reflects any ray from one focus to the other, which is why a whisper at one focus is heard at the other.

Tip

Tip: The focus is the star of every application. Ellipses and hyperbolas have two foci; a parabola has one. That reflecting property is exactly why the focus matters.

Side note
Where it goes wrong

Placing a satellite dish's receiver at the vertex. Signals concentrate at the focus: for a dish y=124x2y = \tfrac{1}{24}x^2 (in inches), 4p=244p = 24, so the receiver goes 66 inches above the vertex.

Try it
Try it: a dish

A satellite dish is 4848 in across and 66 in deep. Where should the receiver be placed?

Show the answer

Model it as y=14px2y = \tfrac{1}{4p}x^2 through (24,6)(24, 6): 6=5764p6 = \tfrac{576}{4p}, so 4p=964p = 96 and p=24p = 24. The receiver goes at the focus, 2424 in above the vertex.

Going Deeper: Advanced Conics

Concept
Building a conic from conditions

So far you have read a conic off its equation. The reverse skill — constructing the equation from geometric conditions — is what real problems ask for. Each unknown coefficient needs one condition:

  • Through given points: plug each point into the general form and solve the resulting linear system for the coefficients.
  • Given foci and eccentricity: the foci fix the center and cc; then e=cae=\dfrac{c}{a} gives aa, and bb comes from b2=a2−c2b^2=a^2-c^2 (ellipse) or b2=c2−a2b^2=c^2-a^2 (hyperbola).
  • Given a tangent line: impose the tangency condition (next box) as one more equation.

Eccentricity ee measures how “stretched” a conic is: e=0e=0 circle, 0<e<10<e<1 ellipse, e=1e=1 parabola, e>1e>1 hyperbola.

Example
Example: ellipse from foci and eccentricity

Find the ellipse centered at the origin with foci (±3,0)(\pm3,0) and eccentricity e=35e=\dfrac35.

The foci are on the xx-axis at distance c=3c=3, so the major axis is horizontal. From e=cae=\dfrac{c}{a},

a=ce=33/5=5,a2=25.a=\frac{c}{e}=\frac{3}{3/5}=5,\qquad a^2=25.

Recompute bb using c2=a2−b2c^2=a^2-b^2:

b2=a2−c2=25−9=16.b^2=a^2-c^2=25-9=16.

So the ellipse is

x225+y216=1,\frac{x^2}{25}+\frac{y^2}{16}=1,

with vertices (±5,0)(\pm5,0), co-vertices (0,±4)(0,\pm4), and foci (±3,0)(\pm3,0) as required.

Concept
The line–conic tangency condition (discriminant =0=0)

To find where a line meets a conic, substitute the line into the conic. You get a quadratic in one variable, so the discriminant Δ=B2−4AC\Delta=B^2-4AC of that quadratic tells you everything:

Δ>0 ⇒ 2 points (secant),Δ=0 ⇒ 1 point (tangent),Δ<0 ⇒ no intersection.\Delta>0 \ \Rightarrow\ 2 \text{ points (secant)},\quad \Delta=0 \ \Rightarrow\ 1 \text{ point (tangent)},\quad \Delta<0 \ \Rightarrow\ \text{no intersection}.

So “the line is tangent to the conic” translates into the single algebraic equation Δ=0\Delta=0. For the ellipse x2a2+y2b2=1\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1 this collapses to the tidy rule: the line y=mx+ky=mx+k is tangent exactly when

k2=a2m2+b2.k^2=a^2m^2+b^2.
Example
Example: tangent lines by the discriminant

For which values of kk is the line y=x+ky=x+k tangent to x24+y2=1\dfrac{x^2}{4}+y^2=1?

Substitute y=x+ky=x+k and clear the fraction (multiply by 44):

x24+(x+k)2=1 ⟹ x2+4(x+k)2=4.\frac{x^2}{4}+(x+k)^2=1 \ \Longrightarrow\ x^2+4(x+k)^2=4.

Expand into a quadratic in xx:

5x2+8kx+(4k2−4)=0.5x^2+8kx+(4k^2-4)=0.

Set its discriminant to zero:

Δ=(8k)2−4(5)(4k2−4)=64k2−80k2+80=−16k2+80=0,\Delta=(8k)^2-4(5)(4k^2-4)=64k^2-80k^2+80=-16k^2+80=0,

so k2=5k^2=5 and k=±5k=\pm\sqrt5. Checking against the shortcut with a2=4a^2=4, b2=1b^2=1, m=1m=1: k2=a2m2+b2=4+1=5k^2=a^2m^2+b^2=4+1=5. ✓

Concept
Latus rectum, focal chords, and the reflection property

A focal chord is any chord through a focus. The special focal chord perpendicular to the major (or focal) axis is the latus rectum, and its length measures how “wide” the curve is at the focus:

ellipse / hyperbola:  2b2a,parabola y=14px2: 4p.\text{ellipse / hyperbola: } \ \frac{2b^2}{a},\qquad \text{parabola } y=\tfrac{1}{4p}x^2: \ 4p.

The reflection property explains every application from Section 7: a ray aimed at one focus of an ellipse bounces to the other focus; a ray parallel to a parabola's axis reflects through its single focus; and a ray toward one focus of a hyperbola reflects away along the line to the other focus.

Example
Example: latus rectum of an ellipse

For x225+y216=1\dfrac{x^2}{25}+\dfrac{y^2}{16}=1 we have a=5a=5, b=4b=4, so each latus rectum has length

2b2a=2(16)5=325=6.4.\frac{2b^2}{a}=\frac{2(16)}{5}=\frac{32}{5}=6.4 .

Since the foci are (±3,0)(\pm3,0), the latus-rectum endpoints above and below the right focus are (3,±165)=(3,±3.2)\left(3,\pm\dfrac{16}{5}\right)=(3,\pm3.2).

Concept
Classifying the full second-degree equation via B2−4ACB^2-4AC

When an xyxy term is present the conic is rotated, and reading AA vs. CC is no longer enough. The general equation

Ax2+Bxy+Cy2+Dx+Ey+F=0Ax^2+Bxy+Cy^2+Dx+Ey+F=0

is classified by the invariant B2−4ACB^2-4AC:

B2−4AC<0 ⇒ ellipse (circle if A=C, B=0),B^2-4AC<0 \ \Rightarrow\ \textbf{ellipse}\ (\text{circle if }A=C,\ B=0),
B2−4AC=0 ⇒ parabola,B2−4AC>0 ⇒ hyperbola.B^2-4AC=0 \ \Rightarrow\ \textbf{parabola},\qquad B^2-4AC>0 \ \Rightarrow\ \textbf{hyperbola}.

This quantity is unchanged by any rotation of the axes, which is why it works even when B≠0B\ne0. (When B=0B=0 it agrees with the sign test of Section 6.)

Example
Example: classify a rotated conic

Classify 2x2+3xy+2y2−5=02x^2+3xy+2y^2-5=0. Here A=2A=2, B=3B=3, C=2C=2, so

B2−4AC=32−4(2)(2)=9−16=−7<0,B^2-4AC=3^2-4(2)(2)=9-16=-7<0,

an ellipse (tilted, since B≠0B\ne0). By contrast x2−4xy+4y2+x=0x^2-4xy+4y^2+x=0 gives B2−4AC=16−4(1)(4)=0B^2-4AC=16-4(1)(4)=0, a parabola.

Concept
Degenerate conics

A “conic” equation does not always draw a smooth curve. When the slicing plane passes through the apex of the cone you get a degenerate conic:

  • A single point — e.g. x2+y2=0x^2+y^2=0 is only (0,0)(0,0) (a “circle of radius 00”).
  • One line / two parallel lines — e.g. x2=4x^2=4 gives x=±2x=\pm2 (a degenerate parabola).
  • Two intersecting lines — e.g. x2−y2=0x^2-y^2=0 gives y=±xy=\pm x (a degenerate hyperbola).
  • Empty set (no points) — e.g. x2+y2=−1x^2+y^2=-1 has no real solutions.

Watch for these after completing the square: a right-hand side of 00 or a negative number is the tell.

Concept
Shortest distance from a point to a conic

The closest point QQ on a conic to an external point PP has a clean geometric signature: the segment PQPQ is perpendicular to the tangent line at QQ (it lies along the normal). For a circle this makes the answer immediate — the nearest point lies on the line through PP and the center:

distance from P to circle=∣ PC‾−r ∣,\text{distance from }P\text{ to circle}=\bigl|\,\overline{PC}-r\,\bigr|,

where CC is the center and rr the radius. For other conics you either minimize the squared distance d2d^2 with calculus or impose the normal condition and solve.

Example
Example: point to a circle

Find the shortest distance from P=(8,0)P=(8,0) to the circle (x−2)2+y2=9(x-2)^2+y^2=9.

The center is C=(2,0)C=(2,0) and r=3r=3. The distance from PP to the center is

PC‾=(8−2)2+(0−0)2=36=6.\overline{PC}=\sqrt{(8-2)^2+(0-0)^2}=\sqrt{36}=6.

Since PP is outside the circle (6>36>3), the shortest distance is

∣PC‾−r∣=∣6−3∣=3,\bigl|\overline{PC}-r\bigr|=|6-3|=3,

reached at the point (5,0)(5,0) where the segment PCPC crosses the circle.

Tip

Tip: Discriminants run this whole section. B2−4ACB^2-4AC (from the equation) tells you which conic; the discriminant of the substituted quadratic (from a line-plus-conic) tells you how a line meets it. Same tool, two jobs.

Try it
Quick check

Write the equation of the circle with center (0,0)(0,0) and radius 66.

Work it on paper before revealing the answer.

Show the answer

Answer: x2+y2=36x^2+y^2=36.

Why It Works

Start from the definition of an ellipse, d1+d2=2ad_1 + d_2 = 2a with foci at (±c,0)(\pm c, 0), write the distances with the distance formula, isolate one radical, square, isolate again, square again, and the equation collapses to x2a2+y2b2=1\tfrac{x^2}{a^2} + \tfrac{y^2}{b^2} = 1 with b2=a2−c2b^2 = a^2 - c^2. The same computation with a minus sign yields the hyperbola. The "focus–directrix" definition of a parabola gives y=14px2y = \tfrac{1}{4p}x^2 in one squaring — which is why a parabolic mirror focuses parallel light at its focus.

Reminder — Distance and midpoint:  d=(x2−x1)2+(y2−y1)2,M=(x1+x22, y1+y22)d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}, \qquad M=\left(\frac{x_1+x_2}{2},\ \frac{y_1+y_2}{2}\right)
Tip
Reading a proof

Do not read a proof the way you read a rule. Cover the last line, ask what the previous line forces, and only then uncover it. If a step feels like magic, that is the step to work by hand with a concrete number.

Formulas, Proofs & Tips

Tip
Equation of a circle
(x−h)2+(y−k)2=r2(x-h)^2+(y-k)^2=r^2

What it means. All points at distance rr from the centre (h,k)(h,k).

Example. Center (2,−1)(2,-1), radius 33: (x−2)2+(y+1)2=9(x-2)^2+(y+1)^2=9.

Why it works. A circle is by definition the set of points a fixed distance from the centre. Writing that distance with the distance formula gives (x−h)2+(y−k)2=r\sqrt{(x-h)^2+(y-k)^2}=r; squaring both sides removes the root.

Tip. Given x2+y2+Dx+Ey+F=0x^2+y^2+Dx+Ey+F=0, complete the square in xx and in yy to recover the centre and radius.