Find standard form of ellipse with vertices (4,0) and (-4,0) And b=1. I know the equation c^2=a^2-b^2 but I can't find a or c. And without a or c I can only plug in b. So can someone help me with this problem and give a good explanation on how I would find a and c so I can figure out how to do the rest of these problems by myself. Please and thank you.
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Find standard form of ellipse
- Thread starter Xmatx5
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Xmatx5 said:Find standard form of ellipse with vertices (4,0) and (4,0) And b=1. I know the equation c^2=a^2-b^2 but I can't find a or c. And without a or c I can only plug in b. So can someone help me with this problem and give a good explanation on how I would find a and c so I can figure out how to do the rest of these problems by myself. Please and thank you.
Please check your problem statement for correctness.
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The vertices are A(4,0) and A'(-4,0) so the midpoint of [AA'] is the center of the ellipse.
This emplyes that 2a=8 so a=4 .
Equation of an ellipse looks like :
(x - 0)^2/a^2 + (y - 0)^2/b^2 = 1
Replace each term by its value, you'll get the equation of the ellipse . . .
Another way , get " c " by using a^2 = b^2 + c^2
MF/MH = e = c / a
The equation of the directrix is a^2/c
And continue . . .
The vertices are A(4,0) and A'(-4,0) so the midpoint of [AA'] is the center of the ellipse.
This emplyes that 2a=8 so a=4 .
Equation of an ellipse looks like :
(x - 0)^2/a^2 + (y - 0)^2/b^2 = 1
Replace each term by its value, you'll get the equation of the ellipse . . .
Another way , get " c " by using a^2 = b^2 + c^2
MF/MH = e = c / a
The equation of the directrix is a^2/c
And continue . . .