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 focus 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
The standard form of an ellipse is given by:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]
To rewrite the given equation \(4x^2 + 9y^2 = 1\) in this form, we divide each term by 1:
\[ \frac{4x^2}{1} + \frac{9y^2}{1} = 1 \]
Now, we can express it as:
\[ \frac{x^2}{\left(\frac{1}{2}\right)^2} + \frac{y^2}{\left(\frac{1}{3}\right)^2} = 1 \]
From this, we can identify:
\[ a^2 = \left(\frac{1}{2}\right)^2 = \frac{1}{4} \quad \text{and} \quad b^2 = \left(\frac{1}{3}\right)^2 = \frac{1}{9} \]

### Step 2: Determine \(a\) and \(b\)
From the above, we have:
\[ a = \frac{1}{2} \quad \text{and} \quad b = \frac{1}{3} \]

### Step 3: Calculate \(c\)
The distance from the center to the foci \(c\) is given by the formula:
\[ c = \sqrt{a^2 - b^2} \]
Substituting the values of \(a^2\) and \(b^2\):
\[ c = \sqrt{\frac{1}{4} - \frac{1}{9}} \]
To perform the subtraction, we need a common denominator:
\[ \frac{1}{4} = \frac{9}{36} \quad \text{and} \quad \frac{1}{9} = \frac{4}{36} \]
Thus,
\[ c = \sqrt{\frac{9}{36} - \frac{4}{36}} = \sqrt{\frac{5}{36}} = \frac{\sqrt{5}}{6} \]

### Step 4: Find the coordinates of the foci
The foci of the ellipse are located at \((\pm c, 0)\). Therefore, 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:
\[ \left(\frac{\sqrt{5}}{6}, 0\right) \quad \text{and} \quad \left(-\frac{\sqrt{5}}{6}, 0\right) \]