Anonymous
Anonymous asked in Science & MathematicsMathematics · 2 months ago

What are the four vertices of (x+2)^2/9 +(y-3)^2/16 =1?

I already found two ((-2,7), (-2,-1)) with the formula but how do I find the other two? 

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8 Answers

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  • 2 months ago

    ((-5, 3), (-2, -1), (-2, 7), (1, 3))

     

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  • 2 months ago

    Let x+2=0, then [(y-3)^2]/16=1=>y-3=+/-4=>

    y=3+4=7 or y=3-4=-1

    => (-2,7) & (-2,-1) are 2 vertices.

    Let y-3=0, then [(x+2)^2]/9=1=>x+2=+/-3=>

    x=-2+3=1 or x-2-3=-5

    => (1,3) & (-5,3) are 2 vertices.

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  • 2 months ago

    You have already found two: (-2,7) and (-2,-1).

    They are √16 = 4 units from their midpoint, which is (-2,3).

    The other two vertices are √9 = 3 units away from this point and lie on the perpendicular bisector of the first two points, so (-1,3) and (5,3).

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  • DWRead
    Lv 7
    2 months ago

    (x+2)²/9 + (y-3)²/16 = 1 is the equation of a vertical ellipse.

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  • 2 months ago

    (x + 2)^2..... (y - 3)^2

    ------------ + -------------- = 1

    ... 9.................. 16

    from this formula 

    (x - h)^2..(y - k)^2

    --------- + ----------- = 1

    ...b^2......... a^2

    finding other vertices of the ellipse

    given our center of ellipse is (- 2,3) and a = 3 and b = 4

    Now use the formula from the coordinates of the minor axis on the vertical ellipse o

    on the left side.

    (h - a,k) = (-2 -3, 3) = (-5, 3) answer//

    (h + a, k) = (-2 + 3, 3) = (1, 3) answer//

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  • sepia
    Lv 7
    2 months ago

    ((x + 2)^2)/9 + ((y - 3)^2)/16 = 1

     ((-2,7), (-2,-1), (-5, 3), (1, 3))

     

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  • david
    Lv 7
    2 months ago

    y  = 3

      ...  ( x+2)^2 = 9

          x + 2  =  +/- 3

           x = -5  and  x = 1

       (-5, 3)  and  (1, 3)

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  • Centered at (-2 , 3)

    Semi-minor axis is 3 units in length and is parallel to the x-axis

    Semi-minor axis is 4 units in length and is parallel to the y-axis

    (-2 - 3 , 3)

    (-2 + 3 , 3)

    (-2 , 3 - 4)

    (-2 , 3 + 4)

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