In the exercises, from 1 to 15, find the standard equation of the hyperbola satisfying the given conditions.
 Foci: \((\pm 7, 0)\). Vertices: \((\pm 5, 0)\).
 Foci: \((\pm 13, 0)\). Vertices: \((\pm 5, 0)\).
 Foci: \((0, \pm6)\). Vertices: \((0, \pm2)\).
 Foci: \((0, \pm 15)\). Vertices: \((0, \pm4)\).
 Foci: (2, 2), (8, 2). Vertices: (0, 2), (6, 2).
 One focus: (3, 3). Vertices: (3, 0), (3, 6 ).
 Foci: (1,2), (5, 2). One vertex: (4, 2).
 Foci: \((\pm 3, 0)\). Asymptotes: \(y = \pm 2x\).
 Foci: \((0,\, \pm\sqrt{58})\). Asymptotes: \(y = \pm \cfrac{5}{2}x\).
 Asymptotes: \(x = \pm \sqrt{3}y\). Passes trough \((6, 4)\).
 Foci: (2, 2), (6, 2), Asymptotes: \(y = x + 2\), \(y = x + 6\).
 Asymptotes: \(y = 2x + 1\), \(y = 2x + 3\). Passes trough (0, 0).
 Vertices: (5, 3), (1, 3). One asymptote: \(2y  x + 7 = 0\).
 Vertices: (1, 3), (3, 3). Eccentricity: \(e=\cfrac{3}{2}\).
 Vertices: (2, 2), (2, 4). Latus rectum: \(L=2\).
In the exercises, from 16 to 19, find the next results completing squares:
 Standard form of the hyperbola.
 The vertices.
 The foci.
 The asymptotes.

\[ 9x^2  16y^2 54x + 64y  127 = 0 \]\[ \begin{aligned} 9x^2  16y^2 54x &+ 64y \\[1em] & 127 = 0 \end{aligned} \]

\[ 4x^2  9y^2 + 32x + 36y + 64 = 0 \]

\[ 16x^2  y^2  32x  6y 57 = 0 \]

\[ 4x^2  9y^2  16x + 54y  101 = 0 \]\[ \begin{aligned} 4x^2  9y^2  16x &+ 54y \\[1em] & 101 = 0 \end{aligned} \]

Two stations \(A\) and \(B\) of a LORAN system are 200\(km\). away on a straight coast.
Station \(B\) is to the west, and station \(A\) is to the east.
One ship receives the signal from the station \(B\), 0.0004 seconds before the signal from the station \(A\).
The speed of the a signal is 300,000\(km/sec\).
One ship receives the signal from the station $B$, 0.0004 seconds before the signal from the station $A$.
The speed of the signal is 300,000\(km/sec\).
 If the ship sails to the coast while keeping a constant time difference, find the equation of the route.
 At what point will the ship reach the coast?
 If the dock is located between the two stations, at 70\(km\). from the station \(B\), find the time difference the ship must keep.
 Two observers stand at the points \(F_1=(200, 0)\) and \(F_2=(200, 0)\) of the plane XY. An explosion at some point of the plane XY is heard by the observer located at \(F_2\) one second before the observer of \(F_1\). Find the equation of the hyperbola where the explosion occurred.
 Find the equation of the set of points $P$ of the plane such that the distance between them and the point (2, 1) is \(\frac{2}{\sqrt{3}}\) the distance between them and the line \(x = \frac{3}{2}\).

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

Prove that the slope of the tangent line to the hyperbola at the point \(P = (x_1, y_1)\) is:
\[ m=\frac{b^2 x_1}{a^2 y_1} \]Hint: The line \(T\): \(y = m(xx_1) + y_1\) pass through \(P = (x_1, y_1)\). \(T\) is tangent to the hyperbola if they intercept at a single point. Follow the same steps given in the solved problem 3.2.6.
Another Hint: Wait until you learn derivatives.

Prove that the a equation of the tangent line \(T\) to the hyperbola at the point \(P=(x_1, y_1)\) is:
\[ \cfrac{x_1 x}{a^2}\cfrac{y_1 y}{b^2}=1 \]

Answers

\[ \frac{x^2}{25} – \frac{y^2}{24} = 1 \]

\[ \frac{x^2}{25} – \frac{y^2}{144} = 1 \]

\[ \frac{y^2}{4} – \frac{x^2}{32} = 1 \]

\[ \frac{y^2}{16} – \frac{x^2}{209} = 1 \]

\[ \frac{ (x – 3)^2 }{9} – \frac{ (y – 2)^2 }{16} = 1 \]

\[ \frac{(y + 3)^2}{9} – \frac{ (x + 3)^2 }{27} = 1 \]

\[ \frac{ (x – 2)^2 }{4} – \frac{(y – 2)^2}{5} = 1 \]

\[ \frac{x^2}{\frac{9}{5}} – \frac{y^2}{\frac{36}{5}} = 1 \]

\[ \frac{y^2}{50} – \frac{x^2}{8} = 1 \]

\[ \frac{y^2}{4} – \frac{x^2}{12} = 1 \]

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

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

\[ \frac{ (y + 2)^2 }{9} – \frac{(x – 3)^2}{36} = 1 \]

\[ \frac{(x – 1)^2}{4} – \frac{(y – 3)^2}{5} = 1 \]

\[ \frac{ (y – 1)^2 }{9} – \frac{(x + 2)^2}{3} = 1 \]


\[ \frac{ (x – 3)^2 }{16} – \frac{(y – 2)^2}{9} = 1 \]

\[ (1, \, 2), \, (7, \, 2) \]

\[ (2, \, 2), \, (8, \, 2) \]

\(3x – 4y – 1 = 0\); \(3x + 4y – 17 = 0\)



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

\[ (4, \, 0), \, (4, \, 4) \]

\(\left( 4, \, 2 – \sqrt{13} \right)\), \(\left( 4, \, 2 + \sqrt{13} \right)\)

\(3y – 2x – 14 = 0\); \(3y + 2x + 2 = 0\)



\[ \frac{(x – 1)^2}{4} – \frac{(y + 3)^2}{64} = 1 \]

\[ (1, \, 3), \, (3, \, 3) \]

\(\left( 1 – 2 \sqrt{17}, \, 3 \right)\), \(\left( 1 + 2\sqrt{17}, \, 3 \right)\)

\(y – 4x + 7 = 0\); \(y + 4x – 1 = 0\)



\[ \frac{(x – 2)^2}{9} – \frac{ (y – 3)^2 }{4} = 1 \]

\[ (1, \, 3), \, (5, \, 3) \]

\[ \left( 2 – \sqrt{13}, \, 3 \right), \, \left( 2 + \sqrt{13}, \, 3 \right) \]

\(2x – 3y + 5 = 0\); \(2x + 3y – 13 = 0\)



\[ \frac{x^2}{3,600} – \frac{y^2}{6,400} = 1 \]

\(40 \, km\) away from the station \(B\).

\(0.0002 \; sec.\)


\[ \frac{x^2}{28,900} – \frac{y^2}{11,100} = 1 \]

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