If the vertex of a hyperbola bisects the distance between its center and the correspoinding focus, then the ratio of the square of its conjugate axis to the square of its transverse axis is (a) 2 (b) 4 (c) 6 (d) 3

Options

A. 2
B. 4
C. 6
D. 3

Explanation

To solve the problem step by step, we need to analyze the given information about the hyperbola and apply the properties of hyperbolas.

### Step 1: Understanding the Hyperbola
The standard form of a hyperbola centered at the origin is given by:
\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]
where \(2a\) is the length of the transverse axis and \(2b\) is the length of the conjugate axis.

### Step 2: Identifying the Focus and Vertex
The foci of the hyperbola are located at \((\pm c, 0)\), where \(c = \sqrt{a^2 + b^2}\). The vertices are at \((\pm a, 0)\).

### Step 3: Given Condition
The problem states that the vertex bisects the distance between the center (which is at the origin \((0, 0)\)) and the corresponding focus. The coordinates of the focus are \((c, 0)\) and the vertex is at \((a, 0)\).

### Step 4: Setting Up the Equation
Since the vertex bisects the distance between the center and the focus, we can express this mathematically:
\[ \text{Midpoint} = \left(\frac{0 + c}{2}, 0\right) = \left(\frac{c}{2}, 0\right) \]
According to the given condition, this midpoint is equal to the vertex \((a, 0)\). Therefore, we have:
\[ \frac{c}{2} = a \implies c = 2a \]

### Step 5: Relating \(c\), \(a\), and \(b\)
From the relationship \(c^2 = a^2 + b^2\), substituting \(c = 2a\) gives:
\[ (2a)^2 = a^2 + b^2 \implies 4a^2 = a^2 + b^2 \]
Simplifying this, we find:
\[ 4a^2 - a^2 = b^2 \implies 3a^2 = b^2 \]

### Step 6: Finding the Ratio of Squares
We need to find the ratio of the square of the conjugate axis to the square of the transverse axis:
\[ \text{Ratio} = \frac{(2b)^2}{(2a)^2} = \frac{4b^2}{4a^2} = \frac{b^2}{a^2} \]
Substituting \(b^2 = 3a^2\) into the ratio gives:
\[ \frac{b^2}{a^2} = \frac{3a^2}{a^2} = 3 \]

### Final Answer
Thus, the ratio of the square of the conjugate axis to the square of the transverse axis is:
\[ \boxed{3} \]