Conics: Part 2: Ellipses and Hyperbolas

HORIZONTAL

$ \displaystyle \begin{array}{c}{{\left( {x-h} \right)}^{2}}+{{\left( {y-k} \right)}^{2}}\\={{r}^{2}}\\\text{Point}\left( {h,k} \right)\text{is center of circle}\end{array}$

$ \displaystyle x=\frac{1}{{4p}}{{\left( {y-k} \right)}^{2}}+h$

or

$ \displaystyle x-h=\frac{1}{{4p}}{{\left( {y-k} \right)}^{2}}$

or

$ 4p\left( {x-h} \right)={{\left( {y-k} \right)}^{2}}$

Example has positive coefficient

Asymptotes:

$ \displaystyle y-k=\pm \frac{b}{a}\left( {x-h} \right)$

VERTICAL

$ \displaystyle y=\frac{1}{{4p}}{{\left( {x-h} \right)}^{2}}+k$

or

$ \displaystyle y-k=\frac{1}{{4p}}{{\left( {x-h} \right)}^{2}}$

 or

$ 4p\left( {y-k} \right)={{\left( {x-h} \right)}^{2}}$

Example has positive coefficient

$ \displaystyle \frac{{{{{\left( {y-k} \right)}}^{2}}}}{{{{a}^{2}}}}+\frac{{{{{\left( {x-h} \right)}}^{2}}}}{{{{b}^{2}}}}=1$

Asymptotes:

$ \displaystyle y-k=\pm \frac{a}{b}\left( {x-h} \right)$

Other

Information

To get $ y$:

$ \begin{array}{c}y=\\\,\,\pm \sqrt{{{{r}^{2}}-{{{\left( {x-h} \right)}}^{2}}}}\\+k\end{array}$

Focal length: $ p$

Focal Width: $ 4p$

Negative Coefficients: Flip parabola

$ {{c}^{2}}={{a}^{2}}-{{b}^{2}}$

Length of Major Axis: $ 2a$

Length of Minor Axis: $ 2b$

$ {{c}^{2}}={{a}^{2}}+{{b}^{2}}$

Length of Transverse Axis: $ 2a$

Length of Conjugate Axis: $ 2b$

 At $ \displaystyle \left( {0,0} \right):\,\,\,\,\frac{{{{x}^{2}}}}{{{{a}^{2}}}}+\frac{{{{y}^{2}}}}{{{{b}^{2}}}}=1$

General:$ \displaystyle \frac{{{{{\left( {x-h} \right)}}^{2}}}}{{{{a}^{2}}}}+\frac{{{{{\left( {y-k} \right)}}^{2}}}}{{{{b}^{2}}}}=1;\,\,\,\,\,a>b$

$ {{a}^{2}}-{{b}^{2}}={{c}^{2}}$

Center: $ \left( {h,k} \right)$     Foci: $ \left( {h\pm c,k} \right)$

Vertices: $ \left( {h\pm a,k} \right)$     Co-Vertices: $ \left( {h,k\pm b} \right)$ 

At $ \displaystyle \left( {0,0} \right):\,\,\,\,\frac{{{{x}^{2}}}}{{{{b}^{2}}}}+\frac{{{{y}^{2}}}}{{{{a}^{2}}}}=1$

General: $ \displaystyle \frac{{{{{\left( {y-k} \right)}}^{2}}}}{{{{a}^{2}}}}+\frac{{{{{\left( {x-h} \right)}}^{2}}}}{{{{b}^{2}}}}=1;\,\,\,\,\,a>b$

$ {{a}^{2}}-{{b}^{2}}={{c}^{2}}$

Center: $ \left( {h,k} \right)$      Foci: $ \left( {h,k\pm c} \right)$

Vertices: $ \displaystyle \left( {h,k\pm a} \right)$     Co-Vertices: $ \left( {h\pm b,k} \right)$  

Notes:  $ a$ is always greater than $ b$;  $ {{a}^{2}}-{{b}^{2}}={{c}^{2}}$;  Major Axis Length $ =2a$;  Minor Axis Length $ =2b$;  Focal Width $ \displaystyle =\frac{{2{{b}^{2}}}}{a}$

Identify the vertices, co-vertices, foci, and domain and range, for the following ellipse, and then graph: $ 9{{x}^{2}}+49{{y}^{2}}=441$.

Domain: $ \left[ {-7,\,7} \right]$    Range: $ \left[ {-3,\,3} \right]$

This would make $ {{a}^{2}}=49$, so $ a=7$. Since the ellipse’s center is $ (0,0)$, the vertices are $ (-7,0)$ and $ (7,0)$. $ {{b}^{2}}=9$, so $ b=3$; the co-vertices are $ (0,-3)$ and $ (0,3)$.

Now let’s find the foci: $ {{c}^{2}}={{a}^{2}}-{{b}^{2}}=49-9=40$. $ c=\sqrt{{40}}\text{ }\left( {\text{or }2\sqrt{{10}}} \right)$, and the foci are $ \displaystyle \left( {\pm \sqrt{{40}},0} \right)$. These can be difficult to graph, but just estimate $ \sqrt{{40}}$ to be close to 6.

Notice that the vertices and foci lie along the horizontal line $ y=0$. The length of the major axis is $ 2a=14$ and the length of the minor axis is $ 2b=6$.

Identify the vertices, co-vertices, foci, and domain and range, for the following ellipse, and then graph: $ \displaystyle \frac{{{{{\left( {x+3} \right)}}^{2}}}}{4}+\frac{{{{{\left( {y-2} \right)}}^{2}}}}{{36}}=1$.

Domain: $ \left[ {-5,-1} \right]$    Range: $ \left[ {-4,\,\,8} \right]$

$ {{a}^{2}}=36$, so $ a=6$. Since the center is $ (–3 ,2)$, the vertices are $ (-3,2-6)$ and $ (-3,2+6)$, or $ (-3,-4)$ and $ (-3,8)$. $ {{b}^{2}}=4$, so $ b=2$; the co-vertices are $ (-3-2,2)$ and $ (-3+2,2)$ or $ (-5,2)$ and $ (-1,2)$.

Now let’s find the foci: $ {{c}^{2}}={{a}^{2}}-{{b}^{2}}=36-4=32$. $ c=\sqrt{{32}}\text{ }\left( {\text{or }4\sqrt{2}} \right)$, and the foci are $ \left( {-3,2\pm \sqrt{{32}}} \right)$. These can be difficult to graph, but just estimate the point: for example, $ 2+\sqrt{{32}}$ is about $ 2+5.5$, or about 7.5.

Notice that the vertices and foci are along the vertical line $ x=-3$. The length of the major axis is $ 2a=12$ and the length of the minor axis is $ 2b=4$.

Identify the vertices, co-vertices, foci and domain and range, for the following ellipse; then graph: $ 4{{x}^{2}}+{{y}^{2}}+24x+2y=-33$.

Domain: $ \left[ {-4,-2} \right]$    Range: $ \left[ {-3,1} \right]$

$ \displaystyle \begin{array}{c}4{{x}^{2}}+24x+{{y}^{2}}+2y=-33\\4\left( {{{x}^{2}}+6x} \right)+{{y}^{2}}+2y=-33\\4\left( {{{x}^{2}}+6x+\,\,\underline{{\,\,\,\,\,\,}}\,} \right)+\left( {{{y}^{2}}+2y+\,\,\underline{{\,\,\,\,\,\,}}\,} \right)=-33+\,\,\underline{{\,\,\,\,\,\,}}\\4\left( {{{x}^{2}}+6x+\color{#2E8B57}{{\underline{{{{{\left( 3 \right)}}^{2}}}}}}} \right)+\left( {{{y}^{2}}+2y+\color{#2E8B57}{{\underline{{{{{\left( 1 \right)}}^{2}}}}}}\,} \right)=-33+\color{#2E8B57}{{\underline{{4{{{\left( 3 \right)}}^{2}}+{{{\left( 1 \right)}}^{2}}}}}}\\4{{\left( {x+\underline{3}} \right)}^{2}}+{{\left( {y+\underline{1}} \right)}^{2}}=-33+36+1\\4{{\left( {x+3} \right)}^{2}}+{{\left( {y+1} \right)}^{2}}=4\end{array}$

$ \displaystyle \frac{{{{{\left( {x+3} \right)}}^{2}}}}{1}+\frac{{{{{\left( {y+1} \right)}}^{2}}}}{4}=1$

We need to first complete the square so we can get the equation in ellipse form. Put all the $ x$’s and $ y$’s together with the constant term on the other side. There can’t be a number (coefficient) before the $ {{x}^{2}}$ or $ {{y}^{2}}$, so factor out the 4 in front of the $ {{x}^{2}}$. Don’t forgot to include it when adding to the right side. After you complete the square, divide all terms by 4, so we have a 1 on the right.

We will use equation: $ \displaystyle \frac{{{{{\left( {y-k} \right)}}^{2}}}}{{{{a}^{2}}}}+\frac{{{{{\left( {x-h} \right)}}^{2}}}}{{{{b}^{2}}}}=1$, since $ 4>1$ (vertical ellipse). We see that the center of the ellipse is at $ (–3,–1)$, so we can first plot that point.

$ {{a}^{2}}=4$, so $ a=2$. Since the center is $ (–3,–1)$, the vertices are $ (–3,–1–2)$ and $ (–3,–1+2)$, or $ (–3,–3)$ and $ (–3,1)$. $ {{b}^{2}}=1$, so $ b=1$. Thus, the co-vertices are $ (-3-1,-1)$ and $ (–3+1,–1)$, or $ (–4,–1)$ and $ (–2,–1)$. Now let’s find the foci: $ {{c}^{2}}={{a}^{2}}-{{b}^{2}}=4-1=3$. $ c=\sqrt{3}$, and the foci are $ \displaystyle \left( {-3,-1\pm \sqrt{3}} \right)$.

Notice that the vertices and foci are along the vertical line $ x=3$. The major axis is $ 2a=4$, and the minor axis is $ 2b=2$.

Domain: $ \left[ {-1-\sqrt{{15}},-1+\sqrt{{15}}} \right]$     Range: $ \left[ {-6,2} \right]$

$ a=4$, so $ {{a}^{2}}=1$. $ c=1$ (distance from center to a focus), so $ {{c}^{2}}=1$. Since $ {{c}^{2}}={{a}^{2}}-{{b}^{2}}$,  $ {{b}^{2}}={{a}^{2}}-{{c}^{2}}=16-1=15$. Thus, $ b=\sqrt{{15}}$, which we’ll need for the domain. Note that the co-vertices would be $ \left( {-1-\sqrt{{15}},-2} \right)$ and $ \left( {-1+\sqrt{{15}},-2} \right)$.

The equation of the ellipse is $ \displaystyle \frac{{{{{\left( {x+1} \right)}}^{2}}}}{{15}}+\frac{{{{{\left( {y+2} \right)}}^{2}}}}{{16}}=1$.

Since the width of the arch is 100 ft, $ a=50$. The height of the arch is 30 feet, so $ b=30$ (the height of the arch is only half of the minor axis of the ellipse). Therefore, we know the equation of the ellipse is $ \displaystyle \frac{{{{x}^{2}}}}{{2500}}+\frac{{{{y}^{2}}}}{{900}}=1$.

We need to get $ c$, so we can find the foci, since the whispering dishes are at the foci. Since $ {{c}^{2}}={{a}^{2}}-{{b}^{2}}$, $ {{c}^{2}}={{50}^{2}}-{{30}^{2}}=1600$, so $ c=40$. Each girl would stand 40 feet from the center of the room.

Now that we have the equation, we can plug in any $ x$ value to get the $ y$ value(s) on the ellipse; since we want the width of the ellipse 15 feet from the vertex, our $ x$ value is $ 75-15=60$. Plugging in 60 for $ x$, we get: $ \displaystyle \frac{{{{{60}}^{2}}}}{{5625}}+\frac{{{{y}^{2}}}}{{1406.25}}=1$; solving for $ y$, we get $ \displaystyle \pm \sqrt{{\left( {1-\frac{{{{{60}}^{2}}}}{{5625}}} \right)\times 1406.25}}=\pm 22.5$ (take positive only). Note that we need to take double 22.5 to get the whole width: the width of the rink 15 feet from a vertex is 45 feet.

 At $ \displaystyle \left( {0,0} \right):\,\,\,\,\,\frac{{{{x}^{2}}}}{{{{a}^{2}}}}-\frac{{{{y}^{2}}}}{{{{b}^{2}}}}=1$

General:  $ \displaystyle \frac{{{{{\left( {x-h} \right)}}^{2}}}}{{{{a}^{2}}}}-\frac{{{{{\left( {y-k} \right)}}^{2}}}}{{{{b}^{2}}}}=1$

$ {{a}^{2}}+{{b}^{2}}={{c}^{2}}$

Center: $ \left( {h,k} \right)$     Foci: $ \left( {h\pm c,k} \right)$  

Vertices: $ \left( {h\pm a,k} \right)$    Co-Vertices:  $ \left( {h,k\pm b} \right)$ 

Length of Transverse Axis: $ 2a$

Length of Conjugate Axis: $ 2b$

Asymptotes: $ \displaystyle y-k=\pm \frac{b}{a}\left( {x-h} \right)$

At $ \displaystyle \left( {0,0} \right):\,\,\,\,\,\frac{{{{y}^{2}}}}{{{{a}^{2}}}}-\frac{{{{x}^{2}}}}{{{{b}^{2}}}}=1$:

General:  $ \displaystyle \frac{{{{{\left( {y-k} \right)}}^{2}}}}{{{{a}^{2}}}}-\frac{{{{{\left( {x-h} \right)}}^{2}}}}{{{{b}^{2}}}}=1$

$ {{a}^{2}}+{{b}^{2}}={{c}^{2}}$

Center: $ \left( {h,k} \right)$     Foci: $ \left( {h,k\pm c} \right)$   

 Vertices: $ \left( {h,k\pm a} \right)$     Co-Vertices: $ \left( {h\pm b,k} \right)$  

Length of Transverse Axis: $ 2a$

Length of Conjugate Axis: $ 2b$

Asymptotes: $ \displaystyle y-k=\pm \frac{a}{b}\left( {x-h} \right)$

Notes: $ {{b}^{2}}$ is always after the minus sign;  $ {{a}^{2}}+{{b}^{2}}={{c}^{2}}$;  Tranverse Axis Length: $ =2a$;  Conjugate Axis Length $ =2b$; Asymptotes are $ \displaystyle y-k=\pm \frac{{\sqrt{{\text{number under the }y}}}}{{\sqrt{{\text{number under the }x}}}}\left( {x-h} \right)$

Identify the center, vertices, foci, and equations of the asymptotes for the following hyperbola; then graph: $ 9{{x}^{2}}-16{{y}^{2}}-144=0$.

Domain: $ \left( {-\infty ,-4} \right]\cup \left[ {4,\infty } \right)$      Range:  $ \left( {-\infty ,\infty } \right)$

Notice that the vertices and foci lie are along the horizontal line $ y=0$, and the length of the transverse axis is $ 2a=8$. The length of the conjugate axis is $ 2b=6$.

This would make $ {{a}^{2}}=16$, so $ a=4$. Since the hyperbola’s center is $ (0,0)$, the vertices are $ (-4,0)$ and $ (4,0)$. $ {{b}^{2}}=9$,  so $ b=3$. Thus, the co-vertices are $ (0,-3)$ and $ (0,3)$. Now we can construct our central rectangle; we use $ a$ and $ b$ to create it.

Now let’s find the foci: $ {{c}^{2}}={{a}^{2}}+{{b}^{2}}=9+16=25$. Thus, $ c=5$ and the foci are $ \displaystyle \left( {\pm \,5,0} \right)$.

The equation of the asymptotes (which go through the corners of the central rectangle) are $ \displaystyle y-k=\pm \frac{{b\text{ (rise)}}}{{a\text{ (run)}}}\left( {x-h} \right)$, or $ \displaystyle y=\pm \frac{3}{4}x$. (Remember to use the square root of what’s under the $ y$ for the numerator of the slope, and the square of what’s under the $ x$ for the denominator.)

Identify the center, vertices, foci, and equations of the asymptotes for the following hyperbola; then graph: $ \displaystyle \frac{{{\left( y+3 \right)}^{2}}}{4}-\frac{{{\left( x-2 \right)}^{2}}}{36}=1$.

Domain: $ \left( {-\infty ,\infty } \right)$       Range: $ \left( {-\infty ,-5} \right]\cup \left[ {-1,\infty } \right)$

Notice that the vertices and foci are along the vertical line $ x=2$, and the length of the transverse axis is $ 2a=4$. The length of the conjugate axis is $ 2b=12$.

$ {{a}^{2}}=4$, so $ a=2$. Since the center is $ (2,–3)$, the vertices are $ (2,–3–2)$ and $ (2,–3+2)$, or $ (2,-5)$ and $ (2,-1)$. $ {{b}^{2}}=36$, so $ b=6$. Thus, the co-vertices are $ (2-6,-3)$ and $ (2+6,-3)$ or $ (-4,-3)$ and $ (8,-3)$.

Now let’s find the foci: $ {{c}^{2}}={{a}^{2}}+\,\,{{b}^{2}}=4+36=40$. Thus, $ c=\sqrt{{40}}\text{ (or }\,2\sqrt{{10}})$, and the foci are $ \displaystyle \left( {2,-3\,\pm \sqrt{{40}}} \right)$.

The equation of the asymptotes (which go through the corners of the central rectangle) are $ \displaystyle y-k=\pm \frac{{a\text{ (rise)}}}{{b\text{ (run)}}}\left( {x-h} \right)$, or $ \displaystyle y+3=\pm \frac{2}{6}\left( {x-2} \right)$ or $ \displaystyle y+3=\pm \frac{1}{3}\left( {x-2} \right)$. (Remember to use the square root of  what’s under the $ y$ for the numerator of the slope, and the square of what’s under the $ x$ for the denominator.)

Identify the center, vertices, foci, and equations of the asymptotes for the following hyperbola; then graph:  $ 49{{y}^{2}}-25{{x}^{2}}+98y-100x+1174=0$.

Domain: $ \left( {-\infty ,-9} \right]\cup \left[ {5,\infty } \right)$    Range: $ \left( {-\infty ,\infty } \right)$

$ \displaystyle \begin{array}{c}49{{y}^{2}}+98y-25{{x}^{2}}-100x=-1174\\49\left( {{{y}^{2}}+2y} \right)-25\left( {{{x}^{2}}+4x} \right)=-1174\\49\left( {{{y}^{2}}+2y+\underline{{\,\,\,\,\,\,}}\,} \right)-25\left( {{{x}^{2}}+4x+\underline{{\,\,\,\,\,\,}}\,} \right)=-1174+\underline{{\,\,\,\,\,\,}}\\49\left( {{{y}^{2}}+2y+\color{#2E8B57}{{\underline{{{{{\left( 1 \right)}}^{2}}}}}}\,} \right)-25\left( {{{x}^{2}}+4x+\color{#2E8B57}{{\underline{{{{{\left( 2 \right)}}^{2}}}}}}\,} \right)=\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-1174+\color{#2E8B57}{{\underline{{49{{{\left( 1 \right)}}^{2}}-25{{{\left( 2 \right)}}^{2}}}}}}\\49{{\left( {y+1} \right)}^{2}}-25{{\left( {x+2} \right)}^{2}}=-1225\end{array}$

$ \displaystyle \frac{{{{{\left( {x+2} \right)}}^{2}}}}{{49}}-\frac{{{{{\left( {y+1} \right)}}^{2}}}}{{25}}=1$

We need to first complete the square so we can get the equation in hyperbola form. Put all the $ x$’s and $ y$’s together with the constant term on the other side. Watch the negative signs; it turns to be a horizontal hyperbola, with the coefficient of the $ {{x}^{2}}$ being positive (we ended up dividing all terms by  –1225).

We will use equation $ \displaystyle \frac{{{{{\left( {x-h} \right)}}^{2}}}}{{{{a}^{2}}}}-\frac{{{{{\left( {y-k} \right)}}^{2}}}}{{{{b}^{2}}}}=1$, since $ x$ comes first (horizontal hyperbola). We see that the center of the hyperbola is at $ (–2,–1)$, so we can first plot that point.

$ {{a}^{2}}=49$, so $ a=7$. Since the center is $ (–2,–1)$, the vertices are $ (–2-7,–1)$ and $ (–2+7,–1)$, or $ (–9,–1)$ and $ (5,–1)$. $ {{b}^{2}}=25$, so $ b=5$. The co-vertices are $ (–2,-1-5)$ and $ (–2,-1+5)$ or $ (–2,–6)$ and $ (–2,4)$.

Now let’s find the foci: $ {{c}^{2}}={{a}^{2}}+{{b}^{2}}=49+25=74$. $ c=\sqrt{{74}}$ , and the foci are $ \displaystyle \left( {-2\pm \sqrt{{74}},\,\,-1\,} \right)$.

The equation of the asymptotes (which go through the corners of the central rectangle) are $ \displaystyle y-k=\pm \frac{{b\text{ (rise)}}}{{a\text{ (run)}}}\left( {x-h} \right)$, or $ \displaystyle y+1=\pm \frac{5}{7}\left( {x+2} \right)$. 

Since the hyperbola is horizontal, we’ll use the equation $ \displaystyle \frac{{{{{\left( {x-h} \right)}}^{2}}}}{{{{a}^{2}}}}-\frac{{{{{\left( {y-k} \right)}}^{2}}}}{{{{b}^{2}}}}=1$.

Plug in our values, and we get $ \displaystyle \frac{{{{{\left( {x-2} \right)}}^{2}}}}{9}-\frac{{{{{\left( {y+5} \right)}}^{2}}}}{9}=1$.

We can see that $ a$ (difference between center and vertex) is 6. So far then we have:

$ \displaystyle \frac{{{{{\left( {x-3} \right)}}^{2}}}}{{{{6}^{2}}}}-\frac{{{{{\left( {y-2} \right)}}^{2}}}}{{{{b}^{2}}}}=\,\,1$

We also see from the asymptotes equation that their slope is $ \displaystyle \pm \frac{2}{3}$. We actually don’t even need to draw them, since we know in our case, since it’s a horizontal hyperbola, we’ll have $ \displaystyle y-2=\pm \frac{{b\text{ (rise)}}}{{6\text{ (run)}}}\left( {x-3} \right)$ (rise is the square root of what’s under the $ y$; run is the square root of what’s under the $ x$) from the equation of the hyperbola above.

Now we can set up a proportion for the asymptote slopes: $ \displaystyle \frac{b}{6}=\frac{2}{3}$; by cross multiplying, we get $ b=4$.

The equation of the hyperbola is: $ \displaystyle \frac{{{{{\left( {x-3} \right)}}^{2}}}}{{{{6}^{2}}}}-\frac{{{{{\left( {y-2} \right)}}^{2}}}}{{{{4}^{2}}}}=1$, or $ \displaystyle \frac{{{{{\left( {x-3} \right)}}^{2}}}}{{36}}-\frac{{{{{\left( {y-2} \right)}}^{2}}}}{{16}}=1$.

(a)  Since the sun is at a focus, we can use the equation $ {{c}^{2}}={{a}^{2}}+{{b}^{2}}$ and take the positive value of $ c$, which is $ \sqrt{{1096+41334}}\approx 205.99$. The coordinates of the sun is $ \left( {0,205.99} \right)$, where each unit is in millions of miles.

(b)  The closest the comet gets the sun as when the comet is at the vertex, which is $ \left( {0,a} \right)$, or $ \left( {0,33.11} \right)$. The closest the comet gets to the sun is about $ 206-33=173$ million miles.

This is actually the “constant difference” of the hyperbola, which is $ 2a$. $ 200-100=2a$, or $ a=50$. Thus, $ {{a}^{2}}=2500$. We also know that $ 2c$ (distance between foci) $ =160$, so $ c=80$. Since $ {{c}^{2}}={{a}^{2}}+{{b}^{2}}$, we can obtain $ {{b}^{2}}:\,{{b}^{2}}={{c}^{2}}-{{a}^{2}}={{80}^{2}}-{{50}^{2}}=3900$.

In the model, the center of the hyperbola is at $ (80,0)$, so the path of the airplane follows the hyperbola$ \displaystyle \frac{{{{{\left( {x-80} \right)}}^{2}}}}{{2500}}-\frac{{{{y}^{2}}}}{{3900}}=1$ .

Now we have to figure out what $ b$ is; we need to use the equation of the asymptote to do this. We do know that $ \displaystyle \frac{b}{a}=\frac{2}{5}$ (formula of asymptotes for horizontal hyperbola), and that $ a=10$. By cross multiplying with $ \displaystyle \frac{b}{{10}}=\frac{2}{5}$, we get $ b=4$.

Therefore, the path of the alpha particle follows the hyperbola $ \displaystyle \frac{{{{x}^{2}}}}{{100}}-\frac{{{{y}^{2}}}}{{16}}=1$.

$ \displaystyle \frac{{{{x}^{2}}}}{{{{a}^{2}}}}-\frac{{{{y}^{2}}}}{{{{b}^{2}}}}=1:\,\,\frac{{{{x}^{2}}}}{{100}}-\frac{{{{y}^{2}}}}{4}=1$

The $ x\text{-}y$ axis can be the surface of the ground, with the plane at $ \left( {0,0} \right)$. The waves of the sonic boom make rings that create a hyperbola intersecting the ground behind the plane.

An observer who hears the sonic boom must be standing on the hyperbola; the closest distance to the plane would be a vertex of the hyperbola; this would be 10 feet away (“$ a$” miles).

  (a) $ 16{{y}^{2}}=-4{{x}^{2}}+64$

  (b) $ 6{{x}^{2}}-6{{y}^{2}}=54$

  (c) $ \displaystyle {{x}^{2}}+{{y}^{2}}=-4x-y+4$

  (d) $ {{x}^{2}}-2y=x+3$

  (e) $ {{y}^{2}}+4y+16{{x}^{2}}+20=0$

(b) The coefficients of $ {{x}^{2}}$ and $ {{y}^{2}}$ are the same,  but we have a $ -$ sign between them, it’s a hyperbola. (We would end up with $ \displaystyle \frac{{{{x}^{2}}}}{9}-\frac{{{{y}^{2}}}}{9}=1$.)

(c) Since the coefficients of $ {{x}^{2}}$ and $ {{y}^{2}}$ are the same, we have a circle. We’d have to complete the square to get it in $ {{\left( {x-h} \right)}^{2}}+{{\left( {y-k} \right)}^{2}}={{r}^{2}}$ form.

(d) Since we have $ {{x}^{2}}$ with other $ y$’s and $ x$’s, we have a parabola. We’d have to complete the square to get it into $ y=a{{\left( {x-h} \right)}^{2}}+k$ (horizontal) form.

(e) Get all variables on one side: $ {{y}^{2}}+4y+16{{x}^{2}}=-20$. Since the coefficients of $ {{x}^{2}}$ and $ {{y}^{2}}$ are different, but we have a  +  sign, it appears to be an ellipse. But after completing the square and dividing by 16, we see we’ll have a $ -1$ on the right-hand side. Thus, it’s none of the conics we’ve studied. (We would end up with $ \displaystyle {{x}^{2}}+\frac{{{{{\left( {y+2} \right)}}^{2}}}}{{16}}=-1$.)

To get $ b$, we can use $ {{c}^{2}}={{a}^{2}}-{{b}^{2}}$, and we know that $ {{c}^{2}}={{\left( {2\sqrt{{13}}} \right)}^{2}}=52$. $ {{b}^{2}}={{a}^{2}}-{{c}^{2}}={{13}^{2}}-52=117$.

The equation of the ellipse is $ \displaystyle \frac{{{{x}^{2}}}}{{117}}+\frac{{{{y}^{2}}}}{{{{{13}}^{2}}}}=1$, or $ \displaystyle \frac{{{{x}^{2}}}}{{117}}+\frac{{{{y}^{2}}}}{{169}}=1$.

We know the center of the hyperbola is $ (-6,-2)$ from the asymptotes. Thus, we have $ \displaystyle \frac{{{{{\left( {y+2} \right)}}^{2}}}}{{{{a}^{2}}}}-\frac{{{{{\left( {x+6} \right)}}^{2}}}}{{{{5}^{2}}}}=1$. To get $ a$, we can use the equation of the asymptotes: we know that  $ \displaystyle \frac{a}{b}$(what’s under the $ y$ over what’s under the $ x$) $ =3$. $ \displaystyle \frac{a}{5}=\frac{3}{1}$, or $ a=15$.

The equation of the hyperbola is $ \displaystyle \frac{{{{{\left( {y+2} \right)}}^{2}}}}{{{{{15}}^{2}}}}-\frac{{{{{\left( {x+6} \right)}}^{2}}}}{{{{5}^{2}}}}=1$.

We also know that $ p$ (distance from vertex to focus) is $ 8-2.5=5.5$, so the equation is $ \displaystyle x=-\frac{1}{{4p}}{{\left( {y-k} \right)}^{2}}+h$, or $ \displaystyle x=-\frac{1}{{22}}{{\left( {y-5} \right)}^{2}}+2.5$.

We have $ \displaystyle \frac{{{{{\left( {x+2} \right)}}^{2}}}}{3}+\frac{{{{{\left( {y-4} \right)}}^{2}}}}{{{{a}^{2}}}}=1$. To get $ a$, we can use $ {{c}^{2}}={{a}^{2}}-{{b}^{2}}$, and, from the foci, we know that $ {{c}^{2}}$ is $ {{\left( {\sqrt{5}} \right)}^{2}}=5$. So, $ {{a}^{2}}={{c}^{2}}+{{b}^{2}}=5+3=8$.

The equation of the ellipse is $ \displaystyle \frac{{{{{\left( {x+2} \right)}}^{2}}}}{3}+\frac{{{{{\left( {y-4} \right)}}^{2}}}}{8}=1$. Tough!