Find the equation of the ellipse with its centre at (4, -1), focus at (1, -1), and passing through (8, 0).

Answer

To find the equation of the ellipse with its center at (4, -1), focus at (1, -1), and passing through the point (8, 0), we can follow these steps:

### Step 1: Identify the center and focus of the ellipse
The center of the ellipse is given as (h, k) = (4, -1). The focus is given as (1, -1).

### Step 2: Determine the orientation of the ellipse
Since the y-coordinates of the center and focus are the same (-1), the ellipse is horizontally oriented.

### Step 3: Write the standard form of the ellipse equation
The standard form of the ellipse equation centered at (h, k) is:
\[ \frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1 \]
Substituting the center (4, -1) into the equation gives:
\[ \frac{(x - 4)^2}{a^2} + \frac{(y + 1)^2}{b^2} = 1 \]

### Step 4: Find the distance from the center to the focus
The distance from the center to the focus (c) can be calculated as:
\[ c = |h - \text{focus}_x| = |4 - 1| = 3 \]
In an ellipse, the relationship between a, b, and c is given by:
\[ c^2 = a^2 - b^2 \]
Thus, we have:
\[ c^2 = 3^2 = 9 \quad \Rightarrow \quad a^2 - b^2 = 9 \]

### Step 5: Use the point (8, 0) to find a and b
Since the ellipse passes through the point (8, 0), we can substitute this point into the ellipse equation:
\[ \frac{(8 - 4)^2}{a^2} + \frac{(0 + 1)^2}{b^2} = 1 \]
This simplifies to:
\[ \frac{4^2}{a^2} + \frac{1^2}{b^2} = 1 \quad \Rightarrow \quad \frac{16}{a^2} + \frac{1}{b^2} = 1 \]

### Step 6: Substitute b² in terms of a²
From the equation \(a^2 - b^2 = 9\), we can express b² as:
\[ b^2 = a^2 - 9 \]
Substituting this into the equation from Step 5 gives:
\[ \frac{16}{a^2} + \frac{1}{a^2 - 9} = 1 \]

### Step 7: Solve for a²
To solve this equation, we can multiply through by \(a^2(a^2 - 9)\) to eliminate the denominators:
\[ 16(a^2 - 9) + a^2 = a^2(a^2 - 9) \]
This simplifies to:
\[ 16a^2 - 144 + a^2 = a^4 - 9a^2 \]
Rearranging gives:
\[ a^4 - 26a^2 + 144 = 0 \]
Letting \(t = a^2\), we have:
\[ t^2 - 26t + 144 = 0 \]

### Step 8: Use the quadratic formula
Using the quadratic formula \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
\[ t = \frac{26 \pm \sqrt{(-26)^2 - 4 \cdot 1 \cdot 144}}{2 \cdot 1} \]
Calculating the discriminant:
\[ t = \frac{26 \pm \sqrt{676 - 576}}{2} = \frac{26 \pm \sqrt{100}}{2} = \frac{26 \pm 10}{2} \]
This gives two possible values:
\[ t = \frac{36}{2} = 18 \quad \text{and} \quad t = \frac{16}{2} = 8 \]

### Step 9: Determine a² and b²
Taking \(t = a^2\):
1. If \(a^2 = 18\), then \(b^2 = 18 - 9 = 9\).
2. If \(a^2 = 8\), then \(b^2 = 8 - 9 = -1\) (not possible).

Thus, we have:
\[ a^2 = 18 \quad \text{and} \quad b^2 = 9 \]

### Step 10: Write the final equation of the ellipse
Substituting \(a^2\) and \(b^2\) back into the ellipse equation gives:
\[ \frac{(x - 4)^2}{18} + \frac{(y + 1)^2}{9} = 1 \]

### Final Answer
The equation of the ellipse is:
\[ \frac{(x - 4)^2}{18} + \frac{(y + 1)^2}{9} = 1 \]