Find the equation of the hyperbola whose vertices are `(pm 6, 0 )` and one of the directices is x = 4 .

Answer

To find the equation of the hyperbola with given vertices and a directrix, we can follow these steps:

### Step 1: Identify the vertices and the standard form of the hyperbola
The vertices of the hyperbola are given as \( (\pm 6, 0) \). This indicates that the hyperbola is centered at the origin and opens horizontally. The standard form of a hyperbola that opens horizontally is:
\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]
where \( a \) is the distance from the center to each vertex.

### Step 2: Determine the value of \( a \)
From the vertices \( (\pm 6, 0) \), we can see that:
\[ a = 6 \]
Thus, \( a^2 = 6^2 = 36 \).

### Step 3: Use the directrix to find \( e \) (eccentricity)
The equation of one of the directrices is given as \( x = 4 \). The distance from the center to the directrix is given by \( \frac{a}{e} \). Since the distance from the center (0,0) to the directrix \( x = 4 \) is 4, we have:
\[ \frac{a}{e} = 4 \]
Substituting the value of \( a \):
\[ \frac{6}{e} = 4 \]
From this, we can solve for \( e \):
\[ e = \frac{6}{4} = \frac{3}{2} \]

### Step 4: Relate \( e \) to \( a \) and \( b \)
The eccentricity \( e \) is related to \( a \) and \( b \) by the formula:
\[ e^2 = 1 + \frac{b^2}{a^2} \]
Substituting \( e = \frac{3}{2} \) and \( a^2 = 36 \):
\[ \left(\frac{3}{2}\right)^2 = 1 + \frac{b^2}{36} \]
Calculating \( \left(\frac{3}{2}\right)^2 \):
\[ \frac{9}{4} = 1 + \frac{b^2}{36} \]
Subtracting 1 from both sides:
\[ \frac{9}{4} - 1 = \frac{b^2}{36} \]
Converting 1 to a fraction:
\[ \frac{9}{4} - \frac{4}{4} = \frac{b^2}{36} \]
This simplifies to:
\[ \frac{5}{4} = \frac{b^2}{36} \]

### Step 5: Solve for \( b^2 \)
Cross-multiplying gives:
\[ 5 \cdot 36 = 4b^2 \]
\[ 180 = 4b^2 \]
Dividing both sides by 4:
\[ b^2 = \frac{180}{4} = 45 \]

### Step 6: Write the equation of the hyperbola
Now that we have \( a^2 = 36 \) and \( b^2 = 45 \), we can write the equation of the hyperbola:
\[ \frac{x^2}{36} - \frac{y^2}{45} = 1 \]

### Final Answer
The equation of the hyperbola is:
\[ \frac{x^2}{36} - \frac{y^2}{45} = 1 \]