Pre Cal.

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1. Identify the conic section represented by: 9y^2+4x^2 - 108y+24x= -144?
Is it ellipse.

2. Find the coordinates of the vertex and the equation of the axis of symmetry for the parabola represented by: x^2 + 4x - 6y + 10 = 0.

vertex: (-2 , 1)
axis of symmetry: x = -2

  • Pre Cal. -

    1.
    9y^2+4x^2 - 108y+24x= -144
    9(y-6)²-9*6² + 4(x+3)²-4*3² = -144
    9(y-6)² + 4(x+3)²= -144 + 324 + 36
    9(y-6)² + 4(x+3)²= (6√6)²
    ((y-6)/(2√6))² + ((x+3)/(3√6))²= 1
    Does that ring a bell?

    2.
    This is done by completing the square:
    x^2 + 4x - 6y + 10 = 0.
    y=(x^2 + 4x + 10)/6
    =(1/6)(x+2)²+1
    =a(x-h)²+k (a=1/6, h=-2, k=1)
    (h,k) is the vertex.
    The equation of the axis of symmetry is
    x=h

  • Pre Cal. -

    identify the conic
    9 y^2+4y^2-108y+24x=-144, determine the center


    i think it is a hyperbola

  • Pre Cal. -

    nvm that last post, i thought this was hw help

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