Identifying Features from Standard Form

Find the center, vertices, foci, and asymptotes of $\frac{(x-2)^2}{16} - \frac{(y+3)^2}{9} = 1$

Identify the form: The $x$ term is positive, so this has a horizontal transverse axis (opens left/right) = Horizontal hyperbola

Find the center $(h, k)$: From $(x-2)^2$ and $(y-(-3))^2$: $h = 2$, $k = -3$ = Center: $(2, -3)$

Find $a$ and $b$: $a^2 = 16$, so $a = 4$; $b^2 = 9$, so $b = 3$ = $a = 4$, $b = 3$

Find the vertices: Vertices are $a$ units left and right of center: $(2 \pm 4, -3)$ = Vertices: $(-2, -3)$ and $(6, -3)$

Find $c$ for the foci: $c = \sqrt{a^2 + b^2} = \sqrt{16 + 9} = \sqrt{25} = 5$ = $c = 5$

Find the foci: Foci are $c$ units from center along transverse axis: $(2 \pm 5, -3)$ = Foci: $(-3, -3)$ and $(7, -3)$

Find the asymptotes: For horizontal: $y - k = \pm\frac{b}{a}(x - h)$, so $y + 3 = \pm\frac{3}{4}(x - 2)$ = Asymptotes: $y = \frac{3}{4}x - \frac{9}{2}$ and $y = -\frac{3}{4}x - \frac{3}{2}$

Answer: Center $(2, -3)$, vertices $(-2, -3)$ and $(6, -3)$, foci $(-3, -3)$ and $(7, -3)$, asymptotes $y + 3 = \pm\frac{3}{4}(x - 2)$

Vertical Hyperbola

Find all features of $\frac{(y-1)^2}{25} - \frac{(x+4)^2}{16} = 1$

Identify the form: The $y$ term is positive, so this has a vertical transverse axis (opens up/down) = Vertical hyperbola

Find the center: From $(y-1)^2$ and $(x-(-4))^2$: $h = -4$, $k = 1$ = Center: $(-4, 1)$

Find $a$ and $b$: $a^2 = 25$ (under $y$), so $a = 5$; $b^2 = 16$, so $b = 4$ = $a = 5$, $b = 4$

Find the vertices: For vertical, vertices are $a$ units above/below center: $(-4, 1 \pm 5)$ = Vertices: $(-4, -4)$ and $(-4, 6)$

Find $c$: $c = \sqrt{25 + 16} = \sqrt{41} \approx 6.4$ = $c = \sqrt{41}$

Find the foci: Foci are $c$ units from center along vertical axis: $(-4, 1 \pm \sqrt{41})$ = Foci: $(-4, 1 - \sqrt{41})$ and $(-4, 1 + \sqrt{41})$

Find the asymptotes: For vertical: $y - k = \pm\frac{a}{b}(x - h)$, so $y - 1 = \pm\frac{5}{4}(x + 4)$ = Asymptotes: $y - 1 = \pm\frac{5}{4}(x + 4)$

Answer: Center $(-4, 1)$, vertices $(-4, -4)$ and $(-4, 6)$, foci $(-4, 1 \pm \sqrt{41})$, asymptotes $y - 1 = \pm\frac{5}{4}(x + 4)$

Writing Standard Form from Features

Write the standard form equation for a hyperbola with center $(0, 0)$, vertex at $(3, 0)$, and focus at $(5, 0)$.

Determine orientation: Vertex and focus are on the $x$-axis, so transverse axis is horizontal = Horizontal hyperbola

Find $a$: Distance from center to vertex: $a = 3$ = $a = 3$, $a^2 = 9$

Find $c$: Distance from center to focus: $c = 5$ = $c = 5$, $c^2 = 25$

Find $b$ using $c^2 = a^2 + b^2$: $25 = 9 + b^2$, so $b^2 = 16$ = $b^2 = 16$

Write the equation: Horizontal form with center at origin: $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ = $\frac{x^2}{9} - \frac{y^2}{16} = 1$

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

Converting to Standard Form

Convert $4x^2 - 9y^2 - 16x - 18y - 29 = 0$ to standard form.

Group $x$ and $y$ terms: $(4x^2 - 16x) - (9y^2 + 18y) = 29$ = Terms grouped

Factor out coefficients: $4(x^2 - 4x) - 9(y^2 + 2y) = 29$ = 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 $y$: $y^2 + 2y + 1 = (y+1)^2$, subtract $9 \cdot 1 = 9$ from right side = $(y+1)^2$

Simplify: $4(x-2)^2 - 9(y+1)^2 = 29 + 16 - 9 = 36$ = Right side: $36$

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

Answer: $\frac{(x-2)^2}{9} - \frac{(y+1)^2}{4} = 1$, center $(2, -1)$, horizontal hyperbola

Mistake: Confusing $a^2$ and $b^2$ positions in horizontal vs vertical forms

Why: In standard form, $a^2$ is always under the POSITIVE term, regardless of whether that's $x$ or $y$.

Correct: Identify which variable has the positive term first. That tells you the orientation, and $a^2$ is under that term.

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

Why: Hyperbolas and ellipses have different relationships between $a$, $b$, and $c$.

Correct: For hyperbolas, $c > a$ and $c^2 = a^2 + b^2$. For ellipses, $c < a$ and $c^2 = a^2 - b^2$.

Mistake: Getting asymptote slopes backwards for vertical hyperbolas

Why: The asymptote formula changes depending on orientation.

Correct: Horizontal: slope $= \pm\frac{b}{a}$. Vertical: slope $= \pm\frac{a}{b}$. Remember: $a$ is always the denominator in the direction the hyperbola opens.

Mistake: Forgetting the $\pm$ when locating vertices and foci

Why: Hyperbolas have two branches, so there are two vertices and two foci.

Correct: Always find BOTH vertices and BOTH foci by adding and subtracting the appropriate distance from the center.

LORAN Navigation

Long Range Navigation uses the time difference of radio signals from two stations. All points with the same time difference form a hyperbola with the stations as foci.

If two LORAN stations are 300 miles apart and a ship receives signals with a time difference corresponding to 100 miles, the ship lies on a hyperbola with $2a = 100$ miles.

Cooling Tower Design

Hyperbolic cooling towers use the shape's structural strength. The hyperbolic profile allows thin shells to support themselves.

A cooling tower has a waist (narrowest point) diameter of 60 meters at height 50 meters, with the hyperboloid equation based on standard form.

Q: How do I remember which form is horizontal vs vertical?

A: Look at which variable has the POSITIVE coefficient. If $x^2$ is positive, the hyperbola opens left/right (horizontal). If $y^2$ is positive, it opens up/down (vertical).

Q: Why is $c^2 = a^2 + b^2$ for hyperbolas but $c^2 = a^2 - b^2$ for ellipses?

A: In an ellipse, the foci are between the vertices ($c < a$). In a hyperbola, the foci are beyond the vertices ($c > a$). The different formulas reflect this geometric relationship.

Q: What do the asymptotes represent?

A: Asymptotes are lines the hyperbola approaches but never touches as it extends to infinity. They help you sketch the hyperbola and define the 'box' that contains the vertices.