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November 23, 2014

November 23, 2014

Posted by **Jeremy** on Monday, August 17, 2009 at 2:15pm.

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. -
**MathMate**, Monday, August 17, 2009 at 6:00pm1.

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. -
**mike**, Tuesday, December 20, 2011 at 9:58pmidentify the conic

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

i think it is a hyperbola

- Pre Cal. -
**mike**, Tuesday, December 20, 2011 at 9:59pmnvm that last post, i thought this was hw help

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