Question
Write the equation of the ellipse centered at the origin with eccentricity e=1/5
and foci at (0,±3/4).
and foci at (0,±3/4).
Answers
GPT-4o mini
To write the equation of the ellipse centered at the origin with the given properties, we need to use the standard form of the equation for an ellipse. The standard form for a vertically oriented ellipse centered at the origin is:
\[
\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1
\]
where:
- \(a\) is the semi-major axis,
- \(b\) is the semi-minor axis,
- \(c\) is the distance from the center to each focus.
We have the following information:
- The foci are at \((0, \pm \frac{3}{4})\), which means \(c = \frac{3}{4}\).
- The eccentricity \(e\) is given as \(e = \frac{1}{5}\).
The relationship between \(a\), \(b\), and \(c\) for an ellipse is given by the equation:
\[
e = \frac{c}{a}
\]
We can rewrite this to find \(a\):
\[
a = \frac{c}{e}
\]
Substituting the known values of \(c\) and \(e\):
\[
a = \frac{\frac{3}{4}}{\frac{1}{5}} = \frac{3}{4} \cdot 5 = \frac{15}{4}
\]
Now we find \(b\) using the relationship \(c^2 = a^2 - b^2\):
First, calculate \(c^2\) and \(a^2\):
\[
c^2 = \left(\frac{3}{4}\right)^2 = \frac{9}{16}
\]
\[
a^2 = \left(\frac{15}{4}\right)^2 = \frac{225}{16}
\]
Now substitute these values into the equation:
\[
\frac{9}{16} = \frac{225}{16} - b^2
\]
Rearranging gives:
\[
b^2 = \frac{225}{16} - \frac{9}{16} = \frac{216}{16} = \frac{27}{2}
\]
Now we have both \(a^2\) and \(b^2\):
- \(a^2 = \frac{225}{16}\)
- \(b^2 = \frac{27}{2}\)
Finally, substituting back into the equation of the ellipse gives:
\[
\frac{x^2}{\frac{27}{2}} + \frac{y^2}{\frac{225}{16}} = 1
\]
To simplify, we can multiply the entire equation by \( \frac{16}{27} \):
\[
\frac{16x^2}{27} + \frac{16y^2}{\frac{225 \cdot 16}{27}} = 1
\]
This can be expressed as:
\[
\frac{16x^2}{27} + \frac{64y^2}{225} = 1
\]
Thus, the equation of the ellipse is:
\[
\frac{16x^2}{27} + \frac{64y^2}{225} = 1
\]
\[
\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1
\]
where:
- \(a\) is the semi-major axis,
- \(b\) is the semi-minor axis,
- \(c\) is the distance from the center to each focus.
We have the following information:
- The foci are at \((0, \pm \frac{3}{4})\), which means \(c = \frac{3}{4}\).
- The eccentricity \(e\) is given as \(e = \frac{1}{5}\).
The relationship between \(a\), \(b\), and \(c\) for an ellipse is given by the equation:
\[
e = \frac{c}{a}
\]
We can rewrite this to find \(a\):
\[
a = \frac{c}{e}
\]
Substituting the known values of \(c\) and \(e\):
\[
a = \frac{\frac{3}{4}}{\frac{1}{5}} = \frac{3}{4} \cdot 5 = \frac{15}{4}
\]
Now we find \(b\) using the relationship \(c^2 = a^2 - b^2\):
First, calculate \(c^2\) and \(a^2\):
\[
c^2 = \left(\frac{3}{4}\right)^2 = \frac{9}{16}
\]
\[
a^2 = \left(\frac{15}{4}\right)^2 = \frac{225}{16}
\]
Now substitute these values into the equation:
\[
\frac{9}{16} = \frac{225}{16} - b^2
\]
Rearranging gives:
\[
b^2 = \frac{225}{16} - \frac{9}{16} = \frac{216}{16} = \frac{27}{2}
\]
Now we have both \(a^2\) and \(b^2\):
- \(a^2 = \frac{225}{16}\)
- \(b^2 = \frac{27}{2}\)
Finally, substituting back into the equation of the ellipse gives:
\[
\frac{x^2}{\frac{27}{2}} + \frac{y^2}{\frac{225}{16}} = 1
\]
To simplify, we can multiply the entire equation by \( \frac{16}{27} \):
\[
\frac{16x^2}{27} + \frac{16y^2}{\frac{225 \cdot 16}{27}} = 1
\]
This can be expressed as:
\[
\frac{16x^2}{27} + \frac{64y^2}{225} = 1
\]
Thus, the equation of the ellipse is:
\[
\frac{16x^2}{27} + \frac{64y^2}{225} = 1
\]