CONIC SECTIONS

Conics — Your Hard Hill

This is the one everyone calls scary. It isn't — it's the same four steps every time. Work through this once and conics goes from "no idea" to "oh, it's just identify-then-plug." ~15–20 min.

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Step 0 every time

Identify First

Look at $x^2$ and $y^2$:

• only ONE squared → parabola

• both, same sign, same number → circle

• both, same sign, different numbers → ellipse

• opposite signs → hyperbola

Then: the positive term names the axis it opens along.

Foci: ellipse $c^2=a^2-b^2$, hyperbola $c^2=a^2+b^2$ (hyperbola adds).

Worked Example 1 — Hyperbola, All the Parts

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

1 · Identify. Two squared terms, opposite signs → hyperbola. The $x$ term is positive → opens left–right.

2 · Center $(h,k)$. Equation has $(x-2)$ and $(y+1)$, so $h=2$, $k=-1$ → center $(2,-1)$. $(y+1)$ means $k=-1$, not $+1$.

3 · $a$ and $b$. $a^2=9 \Rightarrow a=3$ (under the positive term). $b^2=16 \Rightarrow b=4$.

4 · Vertices = center $\pm\, a$ along the open axis ($x$): $(2\pm 3,\,-1)$ → $(5,-1)$ and $(-1,-1)$.

5 · Foci. $c^2=a^2+b^2=9+16=25 \Rightarrow c=5$. Foci = center $\pm\, c$ along $x$: $(7,-1)$ and $(-3,-1)$.

The trap: for a hyperbola, foci use $a^2+b^2$ (add). The ellipse subtracts. If you wrote $9-16$ you'd get a negative — that's your signal you grabbed the ellipse formula.

Worked Example 2 — Completing the Square

The leading-coefficient trap

Put into standard form: $4x^2 + y^2 - 8x - 6y + 9 = 0$

1 · Group $x$'s and $y$'s, move the constant right: $(4x^2-8x)+(y^2-6y)=-9$

2 · Factor the coefficient out of the $x$-group: $4(x^2-2x)+(y^2-6y)=-9$

3 · Complete each square. $x$: half of $-2$ is $-1$, squared $= 1$. $\;y$: half of $-6$ is $-3$, squared $= 9$.

4 · Add to BOTH sides — but the $1$ inside the $\times 4$ bracket adds $4 \cdot 1 = 4$ on the right: $4(x^2-2x+1)+(y^2-6y+9)=-9+4+9$

5 · Compress & simplify: $4(x-1)^2+(y-3)^2=4$

6 · Divide by $4$ to get "$=1$": $\;\dfrac{(x-1)^2}{1}+\dfrac{(y-3)^2}{4}=1$ → ellipse, center $(1,3)$.

Your #1 conic error lives in step 4. The number you add inside a bracket gets multiplied by the coefficient before it lands on the right side. Here the $+1$ became $+4$. Forget the $\times 4$ and every later step is wrong. Write the $\times 4$ out.

Now you

Faded — Fill the Blanks

Same machine as Example 1. Don't peek — fill each blank, then open the answer.

$$\frac{(x+4)^2}{9} - \frac{(y-3)^2}{16} = 1$$

1 · Type: __________ Opens: __________

2 · Center $(\underline{\quad},\underline{\quad})$

3 · $a=\underline{\quad}$ $b=\underline{\quad}$

4 · Vertices $(\underline{\quad},\underline{\quad})$ and $(\underline{\quad},\underline{\quad})$

5 · $c=\underline{\quad}$ Foci $(\underline{\quad},\underline{\quad})$ and $(\underline{\quad},\underline{\quad})$

Check it ›

Type: hyperbola, opens left–right ($x$ positive). Center $(-4,3)$. $a=3$, $b=4$. Vertices $(-1,3)$ and $(-7,3)$. $c=\sqrt{9+16}=5$. Foci $(1,3)$ and $(-9,3)$.

Practice Set (6) — Sheet Face-Down

Do these on paper. Peek at your cheat sheet only when stuck. Then check the key and log any miss.

1 · Identify and write in standard form ›

Problem. $9x^2+25y^2=225$

Divide by $225$ → $\dfrac{x^2}{25}+\dfrac{y^2}{9}=1$. Both positive, different numbers → ellipse, center $(0,0)$.

2 · Identify and state which way it opens ›

Problem. $\dfrac{y^2}{4}-\dfrac{x^2}{9}=1$

Opposite signs → hyperbola; $y^2$ is positive → opens up–down.

3 · Center, vertices, foci (hyperbola) ›

Problem. $\dfrac{(x-1)^2}{16}-\dfrac{(y+2)^2}{9}=1$

Hyperbola. Center $(1,-2)$; $a=4$, $b=3$; opens left–right.

Vertices $(5,-2)$ and $(-3,-2)$.

$c^2=16+9=25 \Rightarrow c=5$ → foci $(6,-2)$ and $(-4,-2)$.

4 · Center, vertices, foci (ellipse) ›

Problem. $\dfrac{(x+2)^2}{4}+\dfrac{(y-1)^2}{25}=1$

Ellipse. Center $(-2,1)$. Bigger denominator ($25$) is under $y$ → major axis vertical, $a=5$, $b=2$.

Vertices $(-2,6)$ and $(-2,-4)$.

$c^2=25-4=21 \Rightarrow c=\sqrt{21}\approx4.6$ → foci $(-2,\,1+\sqrt{21})$ and $(-2,\,1-\sqrt{21})$.

5 · Complete the square, identify ›

Problem. $x^2+y^2+6x-4y-12=0$

Same coefficient on $x^2$ and $y^2$ → circle.

$(x+3)^2+(y-2)^2=12+9+4=25$ → center $(-3,2)$, radius $5$.

6 · Complete the square, identify (with coefficients) ›

Problem. $9x^2+4y^2+18x-16y-11=0$

$9(x^2+2x)+4(y^2-4y)=11$.

Complete: $x\to+1$ (adds $9\cdot1=9$ right), $y\to+4$ (adds $4\cdot4=16$ right).

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

Divide by $36$ → $\dfrac{(x+1)^2}{4}+\dfrac{(y-2)^2}{9}=1$ → ellipse, center $(-1,2)$, major axis vertical ($9$ under $y$).

Before You Leave — Add This to Your Sheet

Write one conic line on your cheat sheet — e.g. "Hyperbola: opposite signs · positive term = the axis · foci $c^2=a^2+b^2$ (ADD)." And log any miss (sign error? wrong foci formula? forgot to multiply by the coefficient?).

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