# Ellipses and Parabolas

1. Find the equation of the parabola with vertex at the origin, that passes through the point (-6,4) and opens upward.

X=1/9y^2

Y=-1/9x^2

X=-1/9y^2

Y=1/9x^2

2. Find the equation for the parabola with the given vertex that passes through the

given point: vertex: (-5,5) point: (-3,17)

y=3/16(x-5)^2+5

y=11/32(x-5)^2+5

y=3(x+5)^2+5

y=11/2(x+5)^2+5

3. Find an equation for the parabola with focus at (-5,0) and vertex at (-5,-4).

x^210x-16y+89=0

x^2+10x+16y+89=0

x+10x-4y+9=0

x^2+10x-16y-39=0

4. Find the standard equation for the ellipse, using either the given characteristics, or characteristics taken from the graph. Vertices: (0, plus or minus 8); foci (0, plus or minus 2sqrt 15)

X^2 y^2

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

4 64

X^2 y^2

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

64 4

X^2 y^2

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

60 64

X^2 y^2

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

64 60

5. Find the eccentricity of the ellipse: x^2 y^2

---- +---- = 1

49 64

6. Identify the equation that represents the graph.

x^2 y^2

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

5 7

x^2 y^2

---- + ---- =1

25 49

x^2 y^2

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

7 5

X^2 y^2

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

49 25

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Ellipses and parabolas are investigated. The solution is detailed and well presented. The response received a rating of "5/5" from the student who originally posted the question.