If `a` hyperbola passes through the foci of the ellipse `(x^2)/(25)+(y^2)/(16)=1` . Its transverse and conjugate axes coincide respectively with the major and minor axes of the ellipse and if the product of eccentricities of hyperbola and ellipse is 1 then the equation of hyperbola is `(x^2)/9-(y^2)/(16)=1` b. the equation of hyperbola is `(x^2)/9-(y^2)/(25)=1` c. focus of hyperbola is (5, 0) d. focus of hyperbola is `(5sqrt(3),0)`

Options

A. the equation of hyperbola is `(x^(2))/(9)-(y^(2))/(16)=1`
B. the equation of the hyperbola is `(x^(2))/(9)-(y^(2))/(25)=1`
C. the vertex of the hyperbola is (5, 0)
D. the vertex of the hyperbola is `(5sqrt3, 0)`

Explanation

To solve the problem step by step, we will follow the logical flow presented in the video transcript.

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

### Step 2: Calculate the eccentricity of the ellipse
The eccentricity \( e \) of the ellipse can be calculated using the formula:
\[ e = \sqrt{1 - \frac{b^2}{a^2}} \]
Substituting the values:
\[ e = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5} \]

### Step 3: Use the relationship between the eccentricities of the hyperbola and ellipse
We are given that the product of the eccentricities of the hyperbola \( (e_1) \) and the ellipse \( (e_2) \) is 1:
\[ e_1 \cdot e_2 = 1 \]
Substituting \( e_2 = \frac{3}{5} \):
\[ e_1 \cdot \frac{3}{5} = 1 \implies e_1 = \frac{5}{3} \]

### Step 4: Establish the relationship between \( a \) and \( b \) for the hyperbola
For a hyperbola, the relationship between the semi-major axis \( a \), semi-minor axis \( b \), and eccentricity \( e \) is given by:
\[ e^2 = 1 + \frac{b^2}{a^2} \]
Substituting \( e_1 = \frac{5}{3} \):
\[ \left(\frac{5}{3}\right)^2 = 1 + \frac{b^2}{a^2} \implies \frac{25}{9} = 1 + \frac{b^2}{a^2} \]
This simplifies to:
\[ \frac{25}{9} - 1 = \frac{b^2}{a^2} \implies \frac{16}{9} = \frac{b^2}{a^2} \implies b^2 = \frac{16}{9} a^2 \]

### Step 5: Write the equation of the hyperbola
The standard form of the hyperbola is:
\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]
Substituting \( b^2 = \frac{16}{9} a^2 \):
\[ \frac{x^2}{a^2} - \frac{y^2}{\frac{16}{9} a^2} = 1 \implies \frac{x^2}{a^2} - \frac{9y^2}{16a^2} = 1 \]
Multiplying through by \( 16a^2 \):
\[ 16x^2 - 9y^2 = 16a^2 \]

### Step 6: Determine the value of \( a^2 \)
We know that the hyperbola passes through the foci of the ellipse, which are located at \( (±5, 0) \) since the foci of the ellipse are given by \( c = \sqrt{a^2 - b^2} = \sqrt{25 - 16} = 3 \). The foci are \( (±c, 0) = (±5, 0) \).

Substituting \( (x, y) = (5, 0) \) into the hyperbola equation:
\[ \frac{5^2}{a^2} - \frac{0}{b^2} = 1 \implies \frac{25}{a^2} = 1 \implies a^2 = 25 \]

### Step 7: Calculate \( b^2 \)
Using \( b^2 = \frac{16}{9} a^2 \):
\[ b^2 = \frac{16}{9} \cdot 25 = \frac{400}{9} \]

### Step 8: Write the final equation of the hyperbola
Now substituting \( a^2 = 25 \) and \( b^2 = \frac{400}{9} \):
\[ \frac{x^2}{25} - \frac{y^2}{\frac{400}{9}} = 1 \implies \frac{x^2}{25} - \frac{9y^2}{400} = 1 \]

### Conclusion
The equation of the hyperbola is:
\[ \frac{x^2}{25} - \frac{9y^2}{400} = 1 \]