The coordinates of a focus of the ellipse `4x^(2) + 9y^(2) =1` are

Options

A. `((sqrt(5))/(6),0)`
B. `(-(sqrt(5))/(6),0)`
C. `((sqrt(3))/(6),0)`
D. `(-(sqrt(3))/(6),0)`

Answer

To find the coordinates of the foci of the ellipse given by the equation \(4x^2 + 9y^2 = 1\), we can follow these steps:

### Step 1: Rewrite the equation in standard form
First, we need to rewrite the equation of the ellipse in standard form. The given equation is:

\[ 4x^2 + 9y^2 = 1 \]

We can divide the entire equation by 1 to get:

\[ \frac{x^2}{\frac{1}{4}} + \frac{y^2}{\frac{1}{9}} = 1 \]

This simplifies to:

\[ \frac{x^2}{\left(\frac{1}{2}\right)^2} + \frac{y^2}{\left(\frac{1}{3}\right)^2} = 1 \]

### Step 2: Identify \(a\) and \(b\)
From the standard form, we can identify \(a\) and \(b\):

- \(a = \frac{1}{2}\) (the semi-major axis)
- \(b = \frac{1}{3}\) (the semi-minor axis)

### Step 3: Determine the eccentricity \(e\)
The eccentricity \(e\) of the ellipse is given by the formula:

\[ e = \sqrt{1 - \frac{b^2}{a^2}} \]

Calculating \(b^2\) and \(a^2\):

- \(b^2 = \left(\frac{1}{3}\right)^2 = \frac{1}{9}\)
- \(a^2 = \left(\frac{1}{2}\right)^2 = \frac{1}{4}\)

Now substituting these values into the eccentricity formula:

\[ e = \sqrt{1 - \frac{\frac{1}{9}}{\frac{1}{4}}} = \sqrt{1 - \frac{4}{9}} = \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3} \]

### Step 4: Calculate the coordinates of the foci
The coordinates of the foci for an ellipse centered at the origin are given by:

\[ (\pm ae, 0) \]

Now we need to calculate \(ae\):

\[ ae = a \cdot e = \frac{1}{2} \cdot \frac{\sqrt{5}}{3} = \frac{\sqrt{5}}{6} \]

Thus, the coordinates of the foci are:

\[ \left(\pm \frac{\sqrt{5}}{6}, 0\right) \]

### Final Answer
The coordinates of the foci of the ellipse \(4x^2 + 9y^2 = 1\) are:

\[ \left(\frac{\sqrt{5}}{6}, 0\right) \text{ and } \left(-\frac{\sqrt{5}}{6}, 0\right) \]

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