Find the equation of the hyperbola whose centre is `(-4,1),` vertex (2,1), and semi-conjugate axis equal to 4.

Answer

To find the equation of the hyperbola with the given parameters, we will follow these steps:

### Step 1: Identify the center, vertex, and semi-conjugate axis
The center of the hyperbola is given as \((-4, 1)\). The vertex is at \((2, 1)\), which indicates that the hyperbola opens horizontally since the y-coordinates of the center and vertex are the same.

### Step 2: Determine the distance from the center to the vertex
The distance from the center \((-4, 1)\) to the vertex \((2, 1)\) is calculated as:
\[ a = |2 - (-4)| = |2 + 4| = 6 \]
Thus, \(a = 6\).

### Step 3: Identify the semi-conjugate axis length
The semi-conjugate axis is given as 4. This means:
\[ b = 4 \]

### Step 4: Write the standard form of the hyperbola
Since the hyperbola opens horizontally, the standard form of the equation is:
\[ \frac{(y - k)^2}{b^2} - \frac{(x - h)^2}{a^2} = 1 \]
where \((h, k)\) is the center of the hyperbola. Here, \(h = -4\) and \(k = 1\).

### Step 5: Substitute the values into the equation
Substituting \(h\), \(k\), \(a\), and \(b\) into the standard form:
\[ \frac{(y - 1)^2}{4^2} - \frac{(x + 4)^2}{6^2} = 1 \]
This simplifies to:
\[ \frac{(y - 1)^2}{16} - \frac{(x + 4)^2}{36} = 1 \]

### Final Equation
Thus, the equation of the hyperbola is:
\[ \frac{(y - 1)^2}{16} - \frac{(x + 4)^2}{36} = 1 \]

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