Exercise 1

Determine and plot the coordinates of the foci and vertices and calculate the eccentricity of the following hyperbolas:

\frac {x^2}{144} - \frac {y^2}{81} = 1

2 \frac {y^2}{144} - \frac {x^2}{25} = 1

2x^2 - 3y^2 = 30

9y^2 - 16x^2 = 1296

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Exercise 2

Determine and plot the coordinates of the foci and vertices and calculate the eccentricity of the following hyperbolas:

4x^2 - 3y^2 - 8x - 8 = 0

y^2 - 2x^2 - 4x - 4y = 0

Exercise 3

Calculate the equation of the hyperbola with a transverse axis of 8 and a focal length of 10.

Exercise 4

The transverse axis of a hyperbola is 12 and the curve passes through the point P = (8, 14). Find its equation.

Exercise 5

Calculate the equation of the hyperbola centered at (0, 0) whose focal length is 34 and the distance from one focus to the closest vertex is 2.

Exercise 6

Determine the equation of the hyperbola centered at (0, 0) that passes through the points: (4, \sqrt{8}) and (2 \sqrt{3}, 2). .

Exercise 7

Determine the equation of the hyperbola centered at (0, 0) that passes through the point (2, \sqrt{3})  and whose eccentricity is \sqrt{3} .

Exercise 8

Determine the equation of the hyperbola centered at (0, 0) knowing that one focus is 2 units from one vertex and 50 from the other.

Exercise 9

Determine the coordinates of the point(s) of intersection between the line x + y − 1 = 0 and the hyperbola x^2 - 2y^2 = 1.

Exercise 10

A rectangular hyperbola passes through the point (4, \frac{1}{2}). Find its equation and determine the coordinates of the vertices and foci.

Exercise 11

The transverse axis of a hyperbola is 12 and the eccentricity is \frac{4}{3}. Calculate the equation of this hyperbola.

Exercise 12

Calculate the equation of a rectangular hyperbola knowing that its focal length is 8 \sqrt{2}.

Exercise 13

The length of the conjugate axis of a hyperbola is 8 and the equations of the asymptotes are: y \pm \frac {2}{3}x. . Calculate the equation of the hyperbola, its foci and vertices.

 

Solution of exercise 1

Determine and plot the coordinates of the foci and vertices and calculate the eccentricity of the following hyperbolas:

1    \frac {x^2}{144} - \frac{y^2}{81} = 1

Determine and plot the coordinates of the foci and vertices

a^2 = 144                  a = 12

b^2 = 81                    b = 9

c = \sqrt{144 + 81}         c = 15

A(12, 0)                    A' (-12, 0)

F(15, 0)                  F'(-15, 0)

e = \frac {15} {12} = \frac {5}{4}

\frac {y^2}{144} - \frac {x^2}{25} = 1

Determine and plot the coordinates of the foci and vertices

a^2 = 144           a = 12

b^2 = 25            b = 5

c = \sqrt{144 + 25}           c = 13

A = (0, 12)              A'= (0, -12)

F= (0, 13)                F' = (0, -13)

\frac{13}{12}

 

3 2x^2 - 3y^2 = 30

Divide by 30:

\frac {x^2}{15} - \frac {y^2 } {10} = 1

a^2 = 15          a = \sqrt{15}

b^2 = 10         b = \sqrt{10}

c = \sqrt{15 + 10}        c = 5

A = (\sqrt{15} , 0)      A' = (- \sqrt{15}, 0)

F = (5, 0)                 F' = (-5, 0)

e = \frac {5} {\sqrt{15}} = \sqrt {15}{3}

4 9y^2 - 16x^2 = 1296

Divide by 1296

Divide by 1296:

\frac{y^2}{144} - \frac {x^2}{81} = 1

a^2 = 144              a = 12

b^2 = 81                b = 9

c = \sqrt{144 + 81}           c = 15

A = (0,12)             A' = (0, -12)

F = (0,15)              F' = (0, -15)

e = \frac {15}{12} = \frac {5}{4}

Solution of exercise 2

Determine and plot the coordinates of the foci and vertices and calculate the eccentricity of the following hyperbolas:

1 4x^2 - 3y^2 - 8x - 8 = 0

exercise solutions

4(x^2 - 2x + 1) - 4 - 3y^2 - 8 = 0

4(x - 1)^2 - 3y^2 = 12

\frac{(x - 1)^2} {3} - \frac {y^2}{4} = 1

a^2 = 3                 a = \sqrt{3}

b^2 = 4                 b = 2

c = \sqrt{3 +4}          c = \sqrt{7}

C = (1,0)

A = (1 + \sqrt{3}, 0)         A' = (1 - \sqrt{3}, 0)

F = (1 + \sqrt{7}, 0)         F' = (1 - \sqrt{7}, 0)

e = \frac {\sqrt{7}}{\sqrt{3}} = \frac {\sqrt {21}} {3}

 

2 y^2 - 2x ^2 - 4x - 4y = 0

exercise solutions

(y^2 - 4y + 4) - 4 - 2(x^2 + 2x + 1) + 2 = 0

(y - 2)^2 - 2 (x + 1)^2 = 2

\frac{(y - 2)^2} {2} - (x + 1)^2 = 1

a^2 = 2             a = \sqrt{2}

b^2 = 1             b = 1

c = \sqrt{2 + 1}          c = \sqrt{3}

C = (-1, 2)

A = (-1, 2 + \sqrt{2})            A' = (-1, 2 - \sqrt{2})

F = (-1, 2 + \sqrt{3})            F' = (-1, 2 - \sqrt{2})

e = \frac{\sqrt{3}} {\sqrt{2}} = \frac {\sqrt{6}} {2}

 

Solution of exercise 3

Calculate the equation of the hyperbola with a transverse axis of 8 and a focal length of 10.

2a = 8             a = 4

2c = 10           c = 5

b = \sqrt{25 - 16}           b = 3

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

 

Solution of exercise 4

The transverse axis of a hyperbola is 12 and the curve passes through the point P = (8, 14). Find its equation.

2a = 12             a = 6

P(8, 14)

\frac{64}{36} - \frac {196} {b^2} = 1

b^2 = 252

\frac{x^2} {36} - \frac{y^2} {252} = 1

 

Solution of exercise 5

Calculate the equation of the hyperbola centered at (0, 0) whose focal length is 34 and the distance from one focus to the closest vertex is 2.

2c = 34                 c = 17

a = 17 - 2 = 15         b = \sqrt{17^2 - 15^2}= 8

\frac{x^2} {225} - \frac{y^2} {64} = 1

 

Solution of exercise 6

Determine the equation of the hyperbola centered at (0, 0) that passes through the points: (4, \sqrt{8}) and (2\sqrt{3}, 2).

\begin{cases} \frac{16}{a^2} - \frac{8}{b^2} = 1 \\ \frac {12}{a^2} - \frac {4}{b^2} = 1 \end{cases}

a^2 = 8

b^2 = 8

\frac{x^2}{8} - \frac{y^2}{8} = 1

x^2 - y^2 = 8

 

Solution of exercise 7

Determine the equation of the hyperbola centered at (0, 0) that passes through the point (2, \sqrt{3}) and whose eccentricity is \sqrt{3}.

\frac{c}{a} = \sqrt{3}             \frac {\sqrt{a^2 + b^2}} {a} = \sqrt{3}

P(2, \sqrt{3})                            \frac{4}{a^2} - \frac {3} {b^2} = 1

\frac {a^2 + b^2}{a^2} = 3         b^2 = 2a^2

\frac{4}{a^2} - \frac {3}{b^2} = 1           \frac{4}{a^2} - \frac {3}{2a^2} = 1

a^2 = \frac {5}{2}                  b^2 = 5

\frac{2x^2}{5}  - \frac {y^2}{5} = 1

 

Solution of exercise 8

Determine the equation of the hyperbola centered at (0, 0) knowing that one focus is 2 units from one vertex and 50 from the other.

2a = 50 - 2 = 48             a = 24

c = 24 + 2 = 26               b = \sqrt{26^2 - 24^2} = 10

\frac{x^2} {576} - \frac {y^2} {100} = 1

 

Solution of exercise 9

Determine the coordinates of the point(s) of intersection between the line x + y − 1 = 0 and the hyperbola x² - 2y²= 1.

 

exercise solutions

\begin{cases}x^2 - 2y^2 = 1 \\ y = 1 - x\end{cases}

x^2 - 4x + 3 = 0

P = (3, -2)

P' = (1, 0)

Solution of exercise 10

A rectangular hyperbola passes through the point (4, 1/2). Find its equation and determine the coordinates of the vertices and foci.

 

exercise solutions

4 \cdot \frac{1} {2} = \frac{a^2}{2}

a^2 = 4

x \cdot y = 2

A = (\sqrt{2}, \sqrt{2})         A' = (-\sqrt{2}, - \sqrt{2})

c = a \sqrt{2} = 2 \sqrt{2} = d(O, F)

F(2,2)             F'(-2, -2)

Solution of exercise 11

The transverse axis of a hyperbola is 12 and the eccentricity is 4/3. Calculate the equation of this hyperbola.

2a = 12              a = 6

\frac{c}{6} = \frac{4}{3}            c = 8

b = \sqrt{64 - 36}

b = \sqrt{28}

\frac{x^2} {36} - \frac{y^2}{28} = 1

 

Solution of exercise 12

Calculate the equation of a rectangular hyperbola knowing that its focal length is 8 \sqrt{2}/ .

2c = 8 \sqrt{2}                c = 4 \sqrt{2}

a = b                                  (4\sqrt{2})^2 = a^2 + a^2

a^2 = 16                           b^2 = 16

\frac{x^2}{16} - \frac{y^2}{16} = 1              x^2 - y^2 = 16

 

Solution of exercise 13

The length of the conjugate axis of a hyperbola is 8 and the equations of the asymptotes are: y = \pm \frac{2}{3}x. Calculate the equation of the hyperbola, its foci and vertices.

2b = 8                b = 4

\frac{2}{3} = \frac{4}{a}              a = 6

c = \sqrt{36 + 16}           c = 2 \sqrt{13}

\frac{x^2}{36} - \frac{y^2}{16} = 1

A = (6, 0)            A' -= (-6, 0)

F = (2 \sqrt{13}, 0)          F' = (-2\sqrt{13}, 0)

 

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Emma

I am passionate about travelling and currently live and work in Paris. I like to spend my time reading, gardening, running, learning languages and exploring new places.