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Math Help - CALC 3 Hyperbolas, Parabolas, and elipse

  1. #1
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    CALC 3 Hyperbolas, Parabolas, and elipse

    Need help with 3 questions.

    1.) A parabola has a vertex at (-2,3) and its directrix is the line y=5. Write an equation of the parabola and find its focus and axis of symmetry.


    • I thought p = 3 , because its the difference between the y coordinate and the directrix.
    • Then I plugged the vertex values for x and y into the equation of a hyperbola like so, (x+2)^2 = 4(-2)(y-3)
    • I then expanded both sides out, x^2 + 4x + 4 = -8y + 24
    • Then I solves for y, y = (-x^2 / 8) - (x / 2) + (5/2)


    2.) write an equation of the hyperbola whose foci are at (4,3) and (4,1) and whose eccentricity is (5/3). Find the vertices, center, the axis, and the asymptotes.


    • I know eccentricity is (c/a) or (c/b) depending on the axis. I'm not sure but I think c = 5 and b = 3
    • Using the equation c^2 = a^2 + b^2 I solve for a ; a = 4
    • Then I think the vertices are (0, 4) ( 0, 5)
    • And I believe the axis is the X-axis
    • Not sure how to get center, and asymptotes


    3.) For the conic 145x^2 + 120xy+ 180y^2 = 900, change the coordinate system using the rotation of axes formulas to the equation of a conic section in standard position. Indicate coordinates (from the original coordinate system) of all foci, vertices, and equations of all axes and asymptotes.


    • Okay I know this is an ellipsecot 2 (phee) = (145 - 180 /120) = ( -7 / 24)
    • 90 degrees < 2(phee) < 180 degrees
    • sin(phee) = sqrt(1-(-7/25) / 2) = (4/5)
    • cos(phee) = sqrt(1+(-7/25) / 2) = (3/5)
    • rotation of axes: x= (3/5)x' + (4/5)y' and y = (4/5)x' + (3/5)y'Not sure how to get foci, vertices, and equations of all axes and asymptotes.
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  2. #2
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    Re: CALC 3 Hyperbolas, Parabolas, and elipse

    1) looks fine. since the directrix is horizontal the axis of symmetry is vertical, through the vertex.

    2) Haven't looked at it.

    3) You have a slip somewhere.

    I have

    x = 4/5 x' + 3/5 y'

    and

    y= -3/5 x' + 4/5 y'

    I lazily subbed these into 145x^2 + 120 xy + 180y^2 =900 using a CAS and it simplified to 100x'^2 + 225 y'^2 = 900.

    This can be further simplified to (x'/3)^2 + (y'/2)^2 = 1.
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