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Math Help - hello, please help me

  1. #1
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    hello, please help me

    Hello everybody, how are you doing

    I have a homework, I already did it almost all, but there are a few exercises which are so hard to me, could you please give me a hand

    one exercise says

    Find the vertices and foci of this ellipse

    x^2/2+y^2/4=1



    another exercise says

    find the center, foci and vertices of the given ellipse

    2x^2+3y^2-8x+6y+5=0


    And the last one, sorry!

    find the equation for the hyperbola described

    Center (0,0) focus at (3,0) vertex at (1,0)

    Thanks for your time and help, and please donít get mad, in other forums people get mad if one asks for help

    Thanks again.
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  2. #2
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    Quote Originally Posted by jhonwashington View Post
    ...
    Find the vertices and foci of this ellipse
    x^2/2+y^2/4=1
    ...
    Hi,

    the axis of the ellipse lie on the coordinate-axis. The center of the ellipse is the origin of the coordinate system. The major axis a is situated on the y-axis, the minor axis b lies on the x-axis.
    the length of the major axis is: l_{a}=\sqrt{4}=2
    the length of the minor axis is: l_{b}=\sqrt{2}

    The foci lie on the y-axis. Let e be the distance of a focus from the center, then you calculate e by:
    e^2=a^2-b^2

    Plug in the values you know. You get F_1(0, \sqrt{2}) \text{ and } F_2(0, -\sqrt{2})

    Oops, sorry, I forgot to calculate the vertices:
    You already know the length of axis. So I think it is no problem for you to get the coordinates of the vertices. For confirmation only: [(0,2),\ (0, -2),\ (\sqrt{2} ,0),\ (-\sqrt{2}, 0)]

    EB
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  3. #3
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    Quote Originally Posted by jhonwashington View Post
    ...
    find the center, foci and vertices of the given ellipse
    2x^2+3y^2-8x+6y+5=0
    ...
    Hi,

    the general equation of an ellipse is:
    \frac{(x-c)^2}{a^2} +\frac{(y-d)^2}{b^2}=1
    where M(c, d) is the center of the ellipse and a and b are the axis of the ellipse.

    You only have to transform the given equation into the general form of the ellipse:

    2x^2+3y^2-8x+6y+5=0
    2(x^2-4x)+3(y^2+2y)=-5. Complete the squares:
    2(x^2-4x+4)+3(y^2+2y+1)=-5+8+3. I've added 4 in the first bracket to get a complete square, so I added 8 at the LHS of the equation. To keep the balance you have to add 8 on the RHS too!
    2(x-2)^2+3(y+1)^2=6. Divide by 6 (Remember the RHS of the general equation is 1)
    \frac{(x-2)^2}{3}+\frac{y+1)^2}{2}=1

    Now you know the length of the axis, the coordinates of the center (M(2,-1)). All further calculations are similar to the first problem. So I leave them to you.

    EB
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  4. #4
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    Quote Originally Posted by jhonwashington View Post
    ...
    find the equation for the hyperbola described
    Center (0,0) focus at (3,0) vertex at (1,0)
    ...
    Hi,

    the general equation of the described hyperbola is:
    \frac{x^2}{a^2}-\frac{y^2}{b^2}=1

    You know by the vertex the length of axis a = 1.

    You know that the coordinate of a focus F(e,0) is calculated by: e^2=a^2+b^2

    Plug in the values you know and you'll get b^2 = 8

    Now screw all parts together and you'll come up with:
    \frac{x^2}{1}-\frac{y^2}{8}=1

    EB
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  5. #5
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    THANKs

    thank you so much for your help earboth
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  6. #6
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    how are you earboth

    I just have a doubt about the exercise 2x^2+3y^2-8x+6y+5=0

    you put as answer to this exercise
    (x-2)^2 + y+1^2 = 1
    _______ _____________
    3 2

    I just don't get how you came up with those denominators 3 and 2? could you please explain me why

    thank you.
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  7. #7
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    Ok I couldn't posted it correctly, I hope you understand what I'm saying

    I tried to copy the answer just as you did, but I couldn't, that line which doesn't have denominator is supposed to have a 2. and the left line just a 3 as denominator.
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  8. #8
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    Quote Originally Posted by jhonwashington View Post
    Ok I couldn't posted it correctly, I hope you understand what I'm saying

    I tried to copy the answer just as you did, but I couldn't, that line which doesn't have denominator is supposed to have a 2. and the left line just a 3 as denominator.
    Quote Originally Posted by earboth View Post
    2(x-2)^2+3(y+1)^2=6
    \frac{2(x-2)^2}{6} + \frac{3(y+1)^2}{6}=1

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

    -Dan
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