The co-ordinates of the foci of `16y^(2) -x^(2) =16` are

Options

A. `(0, pm sqrt(17))`
B. `(pmsqrt(17), 0)`
C. `(0, pm 4)`
D. `(0, pm1)`

Answer

To find the coordinates of the foci of the hyperbola given by the equation \(16y^2 - x^2 = 16\), we can follow these steps:

### Step 1: Rewrite the equation in standard form
Start with the given equation:
\[ 16y^2 - x^2 = 16 \]
Divide both sides by 16 to simplify:
\[ \frac{16y^2}{16} - \frac{x^2}{16} = 1 \]
This simplifies to:
\[ y^2 - \frac{x^2}{16} = 1 \]

### Step 2: Identify the values of \(a^2\) and \(b^2\)
From the standard form of the hyperbola \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\), we can identify:
\[ a^2 = 1 \quad \text{and} \quad b^2 = 16 \]
Thus, we find:
\[ a = 1 \quad \text{and} \quad b = 4 \]

### Step 3: Calculate the eccentricity \(e\)
The eccentricity \(e\) of a hyperbola is given by the formula:
\[ e = \sqrt{1 + \frac{b^2}{a^2}} \]
Substituting the values of \(a^2\) and \(b^2\):
\[ e = \sqrt{1 + \frac{16}{1}} = \sqrt{17} \]

### Step 4: Determine the coordinates of the foci
For a hyperbola of the form \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\), the foci are located at:
\[ (0, \pm ae) \]
Substituting the values of \(a\) and \(e\):
\[ (0, \pm 1 \cdot \sqrt{17}) = (0, \pm \sqrt{17}) \]

### Final Answer
The coordinates of the foci are:
\[ (0, \sqrt{17}) \quad \text{and} \quad (0, -\sqrt{17}) \]

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