If the length of minor axis of the ellipse `(x^(2))/(k^(2)a^(2))+(y^(2))/(b^(2))=1` is equal to the length of transverse axis of hyperbola `(x^(2))/(a^(2))-(y^(2))/(b^(2))=1`,and the equation of ellipse is confocal with hyperbola then the value k is equal to

Options

A. `-sqrt(2)`
B. `sqrt(2)`
C. `-sqrt(3)`
D. `(sqrt(3))`

Answer

To solve the problem, we need to find the value of \( k \) given the conditions about the ellipse and hyperbola. Let's break down the solution step by step.

### Step 1: Understand the given equations
The equation of the ellipse is given by:
\[ \frac{x^2}{k^2 a^2} + \frac{y^2}{b^2} = 1 \]
The equation of the hyperbola is given by:
\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]

### Step 2: Find lengths of axes
- The length of the minor axis of the ellipse is given by \( 2b \).
- The length of the transverse axis of the hyperbola is given by \( 2a \).

### Step 3: Set the lengths equal
According to the problem, the length of the minor axis of the ellipse is equal to the length of the transverse axis of the hyperbola:
\[ 2b = 2a \]
Dividing both sides by 2, we get:
\[ b = a \]

### Step 4: Use the confocal condition
The ellipse is confocal with the hyperbola, which means they share the same foci. The foci of the ellipse and hyperbola can be expressed in terms of their eccentricities.

- The eccentricity \( e_1 \) of the ellipse is given by:
\[ e_1 = \sqrt{1 - \frac{b^2}{k^2 a^2}} \]
- The eccentricity \( e_2 \) of the hyperbola is given by:
\[ e_2 = \sqrt{1 + \frac{b^2}{a^2}} \]

### Step 5: Set foci equal
The foci of the ellipse and hyperbola are given by:
- Foci of the ellipse: \( e_1 \cdot a \)
- Foci of the hyperbola: \( e_2 \cdot a \)

Since they are confocal:
\[ k \cdot e_1 \cdot a = e_2 \cdot a \]
Dividing both sides by \( a \) (assuming \( a \neq 0 \)):
\[ k \cdot e_1 = e_2 \]

### Step 6: Substitute the eccentricities
Substituting the expressions for \( e_1 \) and \( e_2 \):
\[ k \cdot \sqrt{1 - \frac{b^2}{k^2 a^2}} = \sqrt{1 + \frac{b^2}{a^2}} \]

### Step 7: Substitute \( b = a \)
Since we found \( b = a \):
\[ k \cdot \sqrt{1 - \frac{a^2}{k^2 a^2}} = \sqrt{1 + \frac{a^2}{a^2}} \]
This simplifies to:
\[ k \cdot \sqrt{1 - \frac{1}{k^2}} = \sqrt{2} \]

### Step 8: Solve for \( k \)
Squaring both sides:
\[ k^2 \left(1 - \frac{1}{k^2}\right) = 2 \]
This simplifies to:
\[ k^2 - 1 = 2 \]
Thus:
\[ k^2 = 3 \]
Taking the square root:
\[ k = \pm \sqrt{3} \]

### Final Answer
The value of \( k \) is:
\[ k = \sqrt{3} \text{ or } k = -\sqrt{3} \]