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Find the vertices and foci of the elipse. 9x^2+y^2=18

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Answer:

Explanation:

The given equation of the ellipse is in the standard form:

\(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\)

Where \(a\) and \(b\) are the semi-major and semi-minor axes, respectively. In this case, we need to rewrite the given equation in this standard form to find \(a\) and \(b\).

\(9x^2 + y^2 = 18\)

First, divide both sides of the equation by 18 to normalize it:

\(\frac{9x^2}{18} + \frac{y^2}{18} = 1\)

Now, simplify the fractions:

\(\frac{x^2}{2} + \frac{y^2}{18} = 1\)

Comparing this to the standard form equation, we see that \(a^2 = 2\) and \(b^2 = 18\).

To find \(a\) and \(b\), take the square root of both sides:

\(a = \sqrt{2}\)

\(b = \sqrt{18} = 3\sqrt{2}\)

So, the semi-major axis \(a\) is \(\sqrt{2}\) and the semi-minor axis \(b\) is \(3\sqrt{2}\).

Now, we can find the coordinates of the vertices and foci:

The vertices are located at (\(±a, 0\)), so:

Vertices: (\(\sqrt{2}, 0\)) and (-\(\sqrt{2}, 0\))

The distance from the center to each focus is \(c\), where \(c\) can be found using the relationship \(c^2 = a^2 - b^2\):

\(c^2 = (\sqrt{2})^2 - (3\sqrt{2})^2 = 2 - 18 = -16\)

Since \(c\) cannot be negative for a real ellipse, this indicates that the given equation represents an imaginary ellipse, which means there are no real foci for this ellipse.

User NeomerArcana
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