Question 58576: I have a parabolic shape that is shown as: y = 2x^2
I also need to graph an ellipse: (y - 2)^2/1 + x^2/4 = 1
Then, I need to decide if the parabola fits in the hole.
Found 2 solutions by Nate, stanbon:
Answer by Nate(3500)   (Show Source):
Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! I have a parabolic shape that is shown as: y = 2x^2
I also need to graph an ellipse: (y - 2)^2/1 + x^2/4 = 1
Then, I need to decide if the parabola fits in the hole.
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You need to carefully sketch the parabolic shape and the
information about the ellipse on a graph. Then make a
determination regarding the answer to your question.
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Your parabolic shape has a vertex at (0,0) but you haven't
said what it's domain is---that would determine how far up
or down or to he right or to the left it goes.
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The ellipse has its center at (0,2).
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When x=0, (y-2)^2=1 ; y-2=+-1; y=3 or y=1
So the ellipse intersects the y-axis at (0,1) and (0,3)
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When y=2, x^2=4; x=+-2
So the ellipse passes through (2,2) and (-2,2)
When y=0, 4+x^2/4=1; x^2=-12;
So the ellipse does not touch the x-axis.
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Hopefully from all this you can figure out the
answer to your question.
Cheers,
Stan H.
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