Identifying Components from Standard Form

Find the center, vertices, co-vertices, and foci of the ellipse: $\frac{(x-2)^2}{25} + \frac{(y+3)^2}{9} = 1$

Identify the center $(h, k)$: From $(x-2)^2$ and $(y+3)^2$, we have $h = 2$ and $k = -3$ = Center: $(2, -3)$

Find $a^2$ and $b^2$: $a^2 = 25$ (larger), $b^2 = 9$ (smaller) = $a = 5$, $b = 3$

Determine orientation: Since $a^2 = 25$ is under $(x-h)^2$, the major axis is horizontal = Horizontal ellipse

Find vertices (on major axis): $(h \pm a, k) = (2 \pm 5, -3)$ = Vertices: $(-3, -3)$ and $(7, -3)$

Find co-vertices (on minor axis): $(h, k \pm b) = (2, -3 \pm 3)$ = Co-vertices: $(2, -6)$ and $(2, 0)$

Calculate $c$ for foci: $c^2 = a^2 - b^2 = 25 - 9 = 16$, so $c = 4$ = $c = 4$

Find foci (on major axis): $(h \pm c, k) = (2 \pm 4, -3)$ = Foci: $(-2, -3)$ and $(6, -3)$

Answer: Center: $(2, -3)$; Vertices: $(-3, -3)$, $(7, -3)$; Co-vertices: $(2, -6)$, $(2, 0)$; Foci: $(-2, -3)$, $(6, -3)$

Writing Standard Form from Components

Write the standard form equation for an ellipse with center $(0, 0)$, vertices at $(0, \pm 6)$, and foci at $(0, \pm \sqrt{11})$.

Identify orientation: Vertices are on the y-axis, so this is a vertical ellipse = Vertical ellipse

Find $a$: Distance from center to vertex: $a = 6$ = $a = 6$, $a^2 = 36$

Find $c$: Distance from center to focus: $c = \sqrt{11}$ = $c = \sqrt{11}$

Calculate $b$: $c^2 = a^2 - b^2 \Rightarrow 11 = 36 - b^2 \Rightarrow b^2 = 25$ = $b = 5$, $b^2 = 25$

Write the equation: For vertical ellipse centered at origin: $\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1$ = $\frac{x^2}{25} + \frac{y^2}{36} = 1$

Answer: $\frac{x^2}{25} + \frac{y^2}{36} = 1$

Converting to Standard Form

Convert to standard form: $4x^2 + 9y^2 - 16x + 54y + 61 = 0$

Group x and y terms: $(4x^2 - 16x) + (9y^2 + 54y) = -61$ = Terms grouped

Factor coefficients: $4(x^2 - 4x) + 9(y^2 + 6y) = -61$ = Coefficients factored

Complete the square for x: $x^2 - 4x + 4 = (x-2)^2$. Add $4 \cdot 4 = 16$ to right side = $(x-2)^2$

Complete the square for y: $y^2 + 6y + 9 = (y+3)^2$. Add $9 \cdot 9 = 81$ to right side = $(y+3)^2$

Rewrite equation: $4(x-2)^2 + 9(y+3)^2 = -61 + 16 + 81 = 36$ = $4(x-2)^2 + 9(y+3)^2 = 36$

Divide by 36: $\frac{(x-2)^2}{9} + \frac{(y+3)^2}{4} = 1$ = Standard form achieved

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

Mistake: Confusing which denominator gives $a^2$ vs $b^2$

Why: Students assume $a^2$ is always under $x^2$, but $a > b$ by definition, so $a^2$ is the larger denominator.

Correct: Always identify: $a^2$ = larger denominator, $b^2$ = smaller denominator. Then determine orientation based on which variable has $a^2$.

Mistake: Using $c^2 = a^2 + b^2$ (hyperbola formula) instead of $c^2 = a^2 - b^2$

Why: Hyperbolas use addition, ellipses use subtraction. This is a critical difference.

Correct: For ellipses: $c^2 = a^2 - b^2$. Remember: foci are inside the ellipse, so $c < a$.

Mistake: Forgetting to change signs when reading center from equation

Why: The equation has $(x - h)^2$, so $(x + 3)^2$ means $h = -3$, not $h = 3$.

Correct: $(x - h)^2$ with $(x + 3)^2$ means $-h = 3$, so $h = -3$.

Mistake: Placing foci on the minor axis

Why: Students may forget that foci are always on the major axis (the longer one).

Correct: Foci are always on the major axis, at distance $c$ from the center.

Planetary Orbits

All planets orbit the Sun in elliptical paths with the Sun at one focus.

Earth's orbit has $a \approx 149.6$ million km and eccentricity $e \approx 0.017$, making it nearly circular.

Whispering Galleries

In an elliptical room, sound from one focus reflects to the other focus, allowing whispers to travel across the room.

The Capitol Building in Washington D.C. has an elliptical room where this effect occurs.

Q: How do I know if an ellipse is horizontal or vertical?

A: Look at which variable has the larger denominator. If $a^2$ (larger) is under $(x-h)^2$, the major axis is horizontal. If $a^2$ is under $(y-k)^2$, the major axis is vertical.

Q: What is the difference between a and c?

A: $a$ is the distance from center to vertex (on the major axis). $c$ is the distance from center to focus. Since foci are inside the ellipse, $c < a$.

Q: What happens when a = b?

A: When $a = b$, the ellipse becomes a circle! The equation becomes $(x-h)^2 + (y-k)^2 = a^2$, and there are no distinct foci ($c = 0$).