If `e_1` is the eccentricity of the ellipse `x^2/16+y^2/25=1 and e_2` is the eccentricity of the hyperbola passing through the foci of the ellipse and `e_1 e_2=1`, then equation of the hyperbola is

Options

A. the hyperbola is `(x^2)/(9)-(y^2)/(16)=1`
B. the hyperbola is `(x^2)/(9)-(y^2)/(25)=1`
C. a focus of hyperbola is (5,0)
D. a focus of hyperbola is `(5sqrt(3),0)`

Answer

To solve the problem step by step, we will find the eccentricity of the given ellipse, determine the eccentricity of the hyperbola, and then derive the equation of the hyperbola.

### Step 1: Identify the parameters of the ellipse
The equation of the ellipse is given as:
\[ \frac{x^2}{16} + \frac{y^2}{25} = 1 \]
From this equation, we can identify:
- \(a^2 = 16\) (where \(a\) is the semi-major axis)
- \(b^2 = 25\) (where \(b\) is the semi-minor axis)

### Step 2: Calculate the eccentricity of the ellipse
The eccentricity \(e_1\) of an ellipse is given by the formula:
\[ e_1 = \sqrt{1 - \frac{a^2}{b^2}} \]
Substituting the values of \(a^2\) and \(b^2\):
\[ e_1 = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{25 - 16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5} \]

### Step 3: Determine the foci of the ellipse
The foci of the ellipse are located at:
\[ (0, \pm b) = (0, \pm 5) \]

### Step 4: Relate the eccentricity of the hyperbola
We know from the problem that \(e_1 \cdot e_2 = 1\). Thus, we can find \(e_2\):
\[ e_2 = \frac{1}{e_1} = \frac{1}{\frac{3}{5}} = \frac{5}{3} \]

### Step 5: Use the properties of the hyperbola
The eccentricity \(e_2\) of a hyperbola is given by:
\[ e_2 = \sqrt{1 + \frac{b^2}{a^2}} \]
Setting this equal to our calculated \(e_2\):
\[ \frac{5}{3} = \sqrt{1 + \frac{b^2}{a^2}} \]
Squaring both sides:
\[ \left(\frac{5}{3}\right)^2 = 1 + \frac{b^2}{a^2} \]
\[ \frac{25}{9} = 1 + \frac{b^2}{a^2} \]
\[ \frac{25}{9} - 1 = \frac{b^2}{a^2} \]
\[ \frac{25}{9} - \frac{9}{9} = \frac{b^2}{a^2} \]
\[ \frac{16}{9} = \frac{b^2}{a^2} \]
Thus, we have:
\[ b^2 = \frac{16}{9} a^2 \]

### Step 6: Use the foci to find \(b^2\)
The hyperbola passes through the foci of the ellipse, which are at \((0, 5)\) and \((0, -5)\). Substituting \(y = 5\) into the hyperbola equation:
\[ \frac{y^2}{b^2} - \frac{x^2}{a^2} = 1 \]
For \(x = 0\):
\[ \frac{5^2}{b^2} = 1 \implies \frac{25}{b^2} = 1 \implies b^2 = 25 \]

### Step 7: Substitute \(b^2\) back to find \(a^2\)
Using \(b^2 = \frac{16}{9} a^2\):
\[ 25 = \frac{16}{9} a^2 \]
Multiplying both sides by \(9\):
\[ 225 = 16 a^2 \]
\[ a^2 = \frac{225}{16} \]

### Step 8: Write the equation of the hyperbola
The standard form of the hyperbola is:
\[ \frac{y^2}{b^2} - \frac{x^2}{a^2} = 1 \]
Substituting \(b^2 = 25\) and \(a^2 = \frac{225}{16}\):
\[ \frac{y^2}{25} - \frac{x^2}{\frac{225}{16}} = 1 \]
This can be rewritten as:
\[ \frac{y^2}{25} - \frac{16x^2}{225} = 1 \]

### Final Answer
The equation of the hyperbola is:
\[ \frac{y^2}{25} - \frac{16x^2}{225} = 1 \]