Exercise 1

Calculate and plot the coordinates of the foci and vertices and determine the eccentricity of the following ellipses:

1   \frac{x^2}{16} + \frac{y^2} {12} = 1

2   x^2 + 4y^2 = 16

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

4     3x^2 + 2y^2 = 6

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Exercise 2

Calculate and plot the coordinates of the foci and vertices and determine the eccentricity of the following ellipses:

x^2 + 2y^2 - 2x + 8y + 5 = 0

25x^2 + 9y^2 - 18y - 216 = 0

x^2 + 3y^2 - 6x + 6y = 0

3x^2 + y^2 - 24x + 39 = 0

Exercise 3

Determine the equations of the following ellipses using the information given:

C = (0, 0),   F = (2, 0)       A = (3, 0)

C = (0, 0),   F = (0, 4)       A = (0, 5)

C = (1, -1),   F = (1, 2)       A = (1, 4)

C = (-3, 2),   F = (-1, 2)       A = (2, 2)

Exercise 4

Determine the equation of the ellipse that is centered at (0, 0), passes through the point (2, 1) and whose minor axis is 4.

Exercise 5

The focal length of an ellipse is 4 and the distance from a point on the ellipse is 2 and 6 units from each foci respectively. Calculate the equation of the ellipse if it is centered at (0, 0).

Exercise 6

Determine the equation of the ellipse which is centered at (0, 0) and passes through the points:

(1, \frac{\sqrt{3}}{2}) and (\sqrt{2}, \frac{\sqrt{2}}{2}).

Exercise 7

Find the coordinates of the midpoint of the chord in the line: x + 2y − 1 = 0 which intersects the ellipse: x² + 2y² = 3.

Exercise 8

Determine the equation of the ellipse centered at (0, 0) whose focal length is 8\sqrt{6}  and the area of a rectangle in which the ellipse is inscribed within is 80 u².

Exercise 9

Find the equation of the locus of points P (x, y) whose sum of distances to the fixed points (4, 2) and (−2, 2) is equal to 8.

Exercise 10

Determine the equation of the ellipse centered at (0, 0) knowing that one of its vertices is 8 units from a focus and 18 from the other.

Exercise 11

Determine the equation of the ellipse centered at (0, 0) knowing that it passes through the point (0, 4) and its eccentricity is 3/5.

 

Solution of exercise 1

Calculate and plot the coordinates of the foci and vertices and determine the eccentricity of the following ellipses:

\frac{x^2}{16} + \frac{y^2} {12} = 1

a^2 = 16            a = 4                 A = (4, 0)                A' = (-4, 0)

b^2 = 12            b = 2\sqrt{3}          B = (0, 2 \sqrt{3})          B' = (0, -2 \sqrt{3})

c = \sqrt{16 - 12}         c = 2           F(2, 0)             F'(-2, 0)

e = \frac{1}{2}

Ellipse Problems

2   x^2 + 4y^2 = 16

\frac{x^2}{16} + \frac{y^2}{4} = 1

a^2 = 16          a = 4             A = (4, 0)          A' = (-4, 0)

b^2 = 4           b = 2              B = (0, 2)          B' = (0, -2)

c = \sqrt{16 - 4}            c = 2\sqrt{3}         F(2 \sqrt{3}, 0)           F'(-2 \sqrt{3}, 0)

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

 

Ellipse Problems

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

a^2 = 25             a = 5           A = (0, 5)             A' = (0, -5)

b^2 = 9               b = 3            B = (3, 0)           B' = (-3, 0)

c = \sqrt{25 - 9}      c = 4         F(0, 4)            F'(0, -4)

e = \frac{4}{5}

 

Ellipse Problems

4   3x^2 + 2y^2 = 6

\frac{x^2}{2} + \frac{y^2}{3} = 1

a^2 = 3            a = \sqrt{3}          A = (0, \sqrt{3})          A' = (0, -\sqrt{3})

b^2 = 2            b = \sqrt{2}          B = (\sqrt{2}, 0)          B' = (-\sqrt{2}, 0)

c = \sqrt{3 - 2}        c = 1       F(0, 1)            F' (0, -1)

e = \frac{1} {\sqrt{3}} = \frac{\sqrt{3}}{3}

 

Ellipse Problems

 

Solution of exercise 2

Calculate and plot the coordinates of the foci and vertices and determine the eccentricity of the following ellipses:

1    x^2 + 2y^2 - 2x + 8y + 5 = 0

(x^2 - 2x + 1) - 1+ 2(y^2 + 4y + 4) - 8 + 5 = 0

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

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

C = (1, -2)

a^2 = 4         a = 2          A = (3, -2)            A' = (-1, -2)

b^2 = 2         b = \sqrt{2}          B = (1, -2 + \sqrt{2})         B' = (1, -2 - \sqrt{2})

x^2 = \sqrt{4 - 2}              c = \sqrt{2}        F = (1 + \sqrt{2}, -2)        F' = (1 - \sqrt{2}, -2)

 

Ellipse Problems

2     25x^2 + 9y^2 - 18y - 216 = 0

25x^2 + 9(y^2 - 2y + 1) - 9 -216 = 0

25x^2 + 9(y - 1)^2 = 225             \frac{x^2}{9} + \frac{(y - 1)^2} {25} = 1

C = (0, 1)

a^2 = 25                 a = 5              A = (0, 6)           A' = (0, -4)

b^2 = 9                   b = 3             B = (3, 1)            B' = (-3, 1)

C = \sqrt{25 - 9}       c = 4         F(0, 5)       F'(0, -3)

 

Ellipse Problems

3     x^2 + 3y^2 - 6x + 6y = 0

(x^2 -6x + 9) - 9 + 3 (y^2 +2y + 1) - 3 = 0

(x - 3)^2 + 3(y + 1)^2 = 12              \frac{(x - 3)^2}{12} + \frac {(y + 1)^2} {4} = 1

C = (3, -1)

a^2 = 12              a = 2 \sqrt{3}              A = (3 + 2 \sqrt{3}, -1)          A' = (3 - 2\sqrt{3}, -1)

b^2 = 4               b = 2                             B = (3, 1)                 B' = (3, -3)

c = \sqrt {12 - 4}              c = 2 \sqrt{2}            F(3 + 2 \sqrt{2}, -1)            F' (3 - 2\sqrt{2}, -1)

 

Ellipse Problems solutions

4       3x^2 + y^2 - 24x + 39 = 0

3(x^2 - 8x + 16) - 48 + y^2 + 39 = 0

3(x - 4)^2 + y^2 = 9                  \frac{(x - 4)^2} {3} + \frac{y^2}{9} = 1

C = (4, 0)

a^2 = 9                a = 3             A = (4, 3)           A' = (4, -3)

b^2 = 3                b = \sqrt{3}         B = (4 + \sqrt{3}, 0)             B' = (4 - \sqrt{3}, 0)

c = \sqrt{9 -3}           c = \sqrt{6}              F(4, \sqrt{6})          F'(4, -\sqrt{6})

 

Ellipse Problems solutions

Solution of exercise 3

Determine the equations of the following ellipses using the information given:

C = (0, 0),        F = (2, 0),         A = (3, 0)

a = 3          c = 2           b = \sqrt{9 - 4} = \sqrt{5}

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

2    C = (0, 0),           F = (0, 4),         A = (0, 5)

a = 5            c = 4             b = \sqrt{25 - 16} = 3

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

 

3       C = (1, -1),         F = (1, 2)           

    A = (1, 4)<span class="ql-right-eqno">   </span><span class="ql-left-eqno">   </span><img src="https://www.superprof.co.uk/resources/wp-content/ql-cache/quicklatex.com-62e40ae4f0e79dda2246f2f19e9e95cb_l3.png" height="57" width="370" class="ql-img-displayed-equation " alt="\[a = 5$            $c = 3$              $b = \sqrt{25 - 9} = 4\]" title="Rendered by QuickLaTeX.com"/>\frac{(x - 1)^2} {16} + \frac {(y + 1)^2} {25} = 1

4     C = (-3, 2),           F = (-1, 2)          A = (2, 2)

a = 5              c = 2             b = \sqrt{25 - 4} = \sqrt{21}

\frac{(x + 3)^2}{25} + \frac {(y - 2)^2 } {21} = 1

 

Solution of exercise 4

Determine the equation of the ellipse that is centered at (0, 0), passes through the point (2, 1) and whose minor axis is 4.

2b = 4          b = 2

\frac{2^2} {a^2} + \frac{1^2}{2^2} = 1           a = \frac {4}{\sqrt{3}}

\frac{3x^2} {16} + \frac{y^2} {4} = 1

 

Solution of exercise 5

The focal length of an ellipse is 4 and the distance from a point on the ellipse is 2 and 6 units from each foci respectively. Calculate the equation of the ellipse if it is centered at (0, 0).

2c = 4                 c = 2

2a = 2 + 6          a = 4

b = \sqrt{16 - 4} = \sqrt{12}      b = 2 \sqrt{3}

\frac{x^2}{16} + \frac {y^2} {12} = 1

 

Solution of exercise 6

Determine the equation of the ellipse which is centered at (0, 0) and passes through the points:

(1, \frac{\sqrt{3}} {2}) and (\sqrt{2}, \frac{\sqrt{2}} {2})

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

\frac{2} {a^2} + \frac{2}{4b^2} = 1

a = 2     b = 1

\frac{x^2}{4} + y^2 = 1

 

Solution of exercise 7

Find the coordinates of the midpoint of the chord in the line: x + 2y − 1 = 0 which intersects the ellipse: x² + 2y² = 3.

x + 2y - 1 = 0

x^2 + 2y ^2 = 3

A = (-1, 1)

B = (\frac{5}{3}, -\frac{1}{3})

M = (\frac{1 + \frac{5}{3}} {2}, \frac{1 - \frac{1}{3}}{2})

M = (\frac{1}{3}, \frac{1}{3})

 

Ellipse Problems solutions

 

 

 

 

 

Solution of exercise 8

Determine the equation of the ellipse centered at (0, 0) whose focal length is 8 \sqrt{6} and the area of a rectangle in which the ellipse is inscribed within is 80 u².

2c = 8 \sqrt{6}               c = 4 \sqrt{6}

\begin{cases} a^2 = b^2 + (4 \sqrt{6})^2 \\ 2a \cdot 2b = 80\end{cases}

\begin{cases} a^2 = b^2 + 96 \\ b = \frac{20}{a}\end{cases}

a^2 = (\frac{20}{a})^2 + 96              a^4 - 96a^2 - 400 = 0

a = 10               b = 2

\frac{x^2}{100}+ \frac{y^2}{4} = 1

 

Solution of exercise 9

Find the equation of the locus of points P (x, y) whose sum of distances to the fixed points (4, 2) and (−2, 2) is equal to 8.

\overline{PF} + \overline{PF'} = 8

\sqrt{(x + 2)^2 + (y - 2)^2} + \sqrt{(x - 4)^2 + (y - 2)^2} = 8

\sqrt{(x + 2)^2 + (y - 2)^2}  = 8 - \sqrt{(x - 4)^2 + (y - 2)^2 }

3x - 19 = -4\sqrt{(x - 4)^2 + (y - 2)^2}

7x^2 + 16y^2 - 14x - 64y - 41 = 0

 

Solution of exercise 10

Determine the equation of the ellipse centered at (0, 0) knowing that one of its vertices is 8 units from a focus and 18 from the other.

2a = 18 + 8 = 26          a = 13

c = 13 - 8 = 5               b = \sqrt{13^2 - 5 ^2} = 12

\frac{x^2}{169} + \frac{y^2} {144} = 1

 

Solution of exercise 11

Determine the equation of the ellipse centered at (0, 0) knowing that it passes through the point (0, 4) and its eccentricity is 3/5.

\frac{0^2}{a^2} + \frac {4^2}{b^2} = 1            \frac{4^2}{b^2} = 1           b = 4

\frac{3}{5} = \frac{\sqrt{a^2 - 16}}{a}        \frac{9}{25} = \frac{a^2 - 16} {a^2}         9a^2 = 25 a^2 - 400

16a ^2 -400 = 0           a^2 - 25 = 0          a = 5

\frac{x^2} {25} + \frac{y^2}{16} = 1

 

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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.