Find the equation fo the ellipse with its centre at (1, 2) focus at (6, 2) and containing the point (4, 6).

Answer

To find the equation of the ellipse with its center at (1, 2), focus at (6, 2), and containing the point (4, 6), we can follow these steps:

### Step 1: Write the standard form of the ellipse equation
The standard form of the equation of an ellipse centered at (h, k) is given by:

\[ \frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1 \]

Given that the center is (1, 2), we can substitute \(h = 1\) and \(k = 2\):

\[ \frac{(x - 1)^2}{a^2} + \frac{(y - 2)^2}{b^2} = 1 \]

### Step 2: Determine the distance between the center and the focus
The distance \(c\) from the center to the focus can be calculated using the distance formula. The focus is at (6, 2), so:

\[ c = \sqrt{(6 - 1)^2 + (2 - 2)^2} = \sqrt{5^2} = 5 \]

### Step 3: Relate \(a\), \(b\), and \(c\)
For an ellipse, the relationship between \(a\), \(b\), and \(c\) is given by:

\[ c^2 = a^2 - b^2 \]

Substituting \(c = 5\):

\[ 25 = a^2 - b^2 \quad \text{(Equation 1)} \]

### Step 4: Use the point (4, 6) to find another equation
Since the point (4, 6) lies on the ellipse, we can substitute \(x = 4\) and \(y = 6\) into the ellipse equation:

\[ \frac{(4 - 1)^2}{a^2} + \frac{(6 - 2)^2}{b^2} = 1 \]

This simplifies to:

\[ \frac{3^2}{a^2} + \frac{4^2}{b^2} = 1 \]

\[ \frac{9}{a^2} + \frac{16}{b^2} = 1 \quad \text{(Equation 2)} \]

### Step 5: Solve the system of equations
From Equation 1, we have:

\[ a^2 = b^2 + 25 \]

Substituting \(a^2\) into Equation 2:

\[ \frac{9}{b^2 + 25} + \frac{16}{b^2} = 1 \]

Multiplying through by \(b^2(b^2 + 25)\) to eliminate the denominators:

\[ 9b^2 + 16(b^2 + 25) = b^2(b^2 + 25) \]

Expanding and simplifying:

\[ 9b^2 + 16b^2 + 400 = b^4 + 25b^2 \]

\[ 25b^2 + 400 = b^4 + 25b^2 \]

Subtracting \(25b^2\) from both sides:

\[ 400 = b^4 \]

Taking the square root:

\[ b^2 = 20 \]

### Step 6: Find \(a^2\)
Substituting \(b^2 = 20\) back into Equation 1:

\[ a^2 = 20 + 25 = 45 \]

### Step 7: Write the final equation of the ellipse
Now substituting \(a^2\) and \(b^2\) back into the standard form:

\[ \frac{(x - 1)^2}{45} + \frac{(y - 2)^2}{20} = 1 \]

### Final Equation
Thus, the equation of the ellipse is:

\[ \frac{(x - 1)^2}{45} + \frac{(y - 2)^2}{20} = 1 \]