In the exercises, from 1 to 12, find the standard equation of the ellipse satisfying the given conditions.
 Foci: \((\pm 2, 0)\). Vertices: \((\pm 6, 0)\).
 Foci: \((0, \pm 5)\). Vertices: \((0, \pm 8)\).
 Focus: (5, 2). Vertices: (2, 2), (6, 2).
 Focus: (4, 0). Vertices: (4, 2), (4, 8).
 Vertices: (3, 1), (3, 3). Passing through (2, 1).
 Vertices: \((\pm 5, 0)\), Length of the minor axis: 6.
 Vertices: \((1, \pm 3)\). Length of the minor axis: 4.
 Vertices: (2, 10), (2, 10). Passing through (6, 3).
 Focus: \((\pm 2, 0)\). Eccentricity : \(e=2/3\).
 Vertices: (1, 1), (1, 7). Eccentricity : \(e=1/3\).
 Focus: \((\pm 3, 1)\). Length of the latus rectum = 9.
 Center: (2, 2). Vertex: (3, 2). Length of the latus rectum = 4.
In the exercises, from 13 to 15, the graph of the given equation is an ellipse. Find:
 completing squares, the standard equation.
 the vertices and the foci.

\[ x^2 + 4y^2  6x + 16y =  9 \]

\[ 9x^2 + 4y^2 18x + 24y + 9 = 0 \]

\[ 25x^2 + 16y^2 +100x  96y = 156 \]

A semielliptical archway over a twoway road has a height of 10 feet and a width of 30 feet.
A truck is approaching on its side of the road. It has a width of 9 feet and a height of 7.5 feet.
Can the the truck cross the tunnel without hitting the archway and not using the other side of the road?

The section through the longest part of the Montreal Olympics Stadium is an ellipse. The length of the major and minor axes are 480 \(m\). and 280 \(m\). respectively. Find the standard equation of this ellipse.
 Let \(S\) be the set of points \(P\) of the plane such that the distance from \(P\) to the point (2, 0) is a half the distance from \(P\) to the line \(x = 8\). Find an equation that satisfy all the points of \(S\). Identify this set.
 Let \(S\) be the set of points \(P\) of the plane such that the distance from \(P\) to the point (0, 4) is \(\frac{4}{5}\) the distance from \(P\) to the line \(y = \frac{25}{4}\). Find equation that satisfy all points of \(S\). Identify this set.

A satellite has an elliptical orbit where the center of the moon is at one of the foci. The maximum and minimum distances from the satellite to the moon surface are 4,522 and 522 \(km\), respectively. The radius of the moon is 1,728 \(Km\). Find:
 the eccentricity of the orbit.
 the standard equation of the orbit using the coordinate system with origin at the center of the orbit, and the center of the moon on the Xaxis positive side.

Let \(P = (x_1, y_1)\) be a point of the ellipse \(\cfrac{x^2}{a^2}+ \cfrac{y^2}{b^2}=1\);

Prove that the slope of the tangent line to the ellipse at the point \(P = (x_1, y_1)\) is:
\[ m = \frac{b^2x_1}{a^2y_1} \]First hint: The line \(T\): \(y = m(x x_1) + y_1\) pass through \(P = (x_1, y_1)\).
Second hint: \(T\) is tangent to the ellipse if \(T\) intersects the ellipse in a unique point. Follow the steps of the solved problem 3.2.6.
Third hint: Wait until you learn derivatives.
 Prove that an equation of the tangent line \(T\) to an ellipse at the point \(P = (x_1, y_1)\) is the following: \[ \cfrac{x_1 x}{a^2} + \cfrac{y_1 y}{b^2}=1 \]

Answers

\[ \frac{x^2}{36} + \frac{y^2}{32} = 1 \]

\[ \frac{x^2}{39} + \frac{y^2}{64} = 1 \]

\[ \frac{(x – 4)^2}{4} + \frac{(y – 2)^2}{3} = 1 \]

\[ \frac{(x + 4)^2}{16} + \frac{(y – 3)^2}{25} = 1 \]

\[ (x – 3)^2 + \frac{( y – 1 )^2 }{4} = 1 \]

\[ \frac{ x^2 }{25} + \frac{y^2}{9} = 1 \]

\[ \frac{ (x + 1)^2 }{4} + \frac{y^2}{9} = 1 \]

\[ \frac{ (x – 2)^2 }{ \frac{1600}{91} } + \frac{y^2}{100} = 1 \]

\[ \frac{ x^2 }{9} + \frac{y^2}{5} = 1 \]

\[ \frac{(x + 1)^2}{8} + \frac{(y + 4)^2}{9} = 1 \]

\[ \frac{x^2}{36} + \frac{(y + 1)^2}{27} = 1 \]


\(\cfrac{(x – 3)^2}{4} + \cfrac{(y + 2)^2}{1} = 1\)

Vertices: \((1, \, 2)\), \((5, \, 2)\). Foci: \(\left( 3 – \sqrt{3}, \, 2 \right)\), \(\left( 3 + \sqrt{3}, \, 2 \right)\)



\(\cfrac{ (x – 1)^2 }{4} + \cfrac{(y + 3)^2}{9} = 1\)

Vertices: \((1, \, 6)\), \((1, \, 0)\). Foci: \(\left( 1, \, 3 \sqrt{5} \right)\), \(\left(1, \, 3 + \sqrt{5} \right)\)



\(\cfrac{(x + 2)^2}{16} + \cfrac{(y – 3)^2}{25} = 1\)

Vertices: \((2, \, 2)\), \((2, \, 8)\). Foci: \((2, \, 0)\), \((2, \, 6)\)


Yes. The height of the tunnel when \(x = 9\) is \(8\), and \(7.5 < 8\).

\[ \frac{ x^2 }{(240)^2} + \frac{y^2}{(140)^2} = 1 \]

\[ \frac{x^2}{16} + \frac{y^2}{12} = 1. \; \text{ Ellipse}. \]

\[ \frac{x^2}{9} + \frac{y^2}{25} = 1. \; \text{ Ellipse}. \]


\[ e = \frac{8}{17} \]

\[ \frac{x^2}{ 18,062,500 } + \frac{ y^2 }{ 14,062,500 } = 1 \]\[ \begin{aligned} \frac{x^2}{ 18,062,500 } &+ \frac{ y^2 }{ 14,062,500 } \\[.5em] &\hspace{3em} = 1 \end{aligned} \]
