Find the equation of the ellipse with axes along the x-axis and the y-axis, which passes through the points P(4, 3) and Q(6, 2).

Answer

To find the equation of the ellipse with axes along the x-axis and the y-axis that passes through the points P(4, 3) and Q(6, 2), we can follow these steps:

### Step 1: Write the standard form of the ellipse
The standard form of the ellipse with axes along the x-axis and y-axis is given by:

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

where \(a\) is the semi-major axis and \(b\) is the semi-minor axis.

### Step 2: Substitute the first point P(4, 3) into the equation
Substituting the coordinates of point P(4, 3) into the equation:

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

This simplifies to:

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

### Step 3: Rearrange the equation
Let \(p = \frac{1}{a^2}\) and \(q = \frac{1}{b^2}\). The equation becomes:

\[ 16p + 9q = 1 \quad \text{(Equation 1)} \]

### Step 4: Substitute the second point Q(6, 2) into the equation
Now, substituting the coordinates of point Q(6, 2) into the equation:

\[ \frac{6^2}{a^2} + \frac{2^2}{b^2} = 1 \]

This simplifies to:

\[ \frac{36}{a^2} + \frac{4}{b^2} = 1 \]

### Step 5: Rearrange the second equation
Using the same substitutions \(p\) and \(q\):

\[ 36p + 4q = 1 \quad \text{(Equation 2)} \]

### Step 6: Solve the system of equations
Now we have a system of equations:

1. \(16p + 9q = 1\)
2. \(36p + 4q = 1\)

We can solve these equations using the elimination method.

### Step 7: Multiply the equations to eliminate \(q\)
To eliminate \(q\), we can multiply Equation 1 by 4 and Equation 2 by 9:

\[ 4(16p + 9q) = 4 \quad \Rightarrow \quad 64p + 36q = 4 \quad \text{(Equation 3)} \]

\[ 9(36p + 4q) = 9 \quad \Rightarrow \quad 324p + 36q = 9 \quad \text{(Equation 4)} \]

### Step 8: Subtract Equation 3 from Equation 4
Now, subtract Equation 3 from Equation 4:

\[ (324p + 36q) - (64p + 36q) = 9 - 4 \]

This simplifies to:

\[ 260p = 5 \]

### Step 9: Solve for \(p\)
Dividing both sides by 260 gives:

\[ p = \frac{5}{260} = \frac{1}{52} \]

### Step 10: Substitute \(p\) back to find \(q\)
Now substitute \(p\) back into Equation 1 to find \(q\):

\[ 16\left(\frac{1}{52}\right) + 9q = 1 \]

This simplifies to:

\[ \frac{16}{52} + 9q = 1 \quad \Rightarrow \quad 9q = 1 - \frac{16}{52} \]

Calculating the right side:

\[ 1 - \frac{16}{52} = \frac{52 - 16}{52} = \frac{36}{52} = \frac{9}{13} \]

Thus:

\[ 9q = \frac{9}{13} \quad \Rightarrow \quad q = \frac{1}{13} \]

### Step 11: Find \(a^2\) and \(b^2\)
Now we can find \(a^2\) and \(b^2\):

\[ a^2 = \frac{1}{p} = 52, \quad b^2 = \frac{1}{q} = 13 \]

### Step 12: Write the final equation of the ellipse
Substituting \(a^2\) and \(b^2\) back into the standard form of the ellipse:

\[ \frac{x^2}{52} + \frac{y^2}{13} = 1 \]

### Step 13: Multiply through by 13 to simplify
To express it in a more standard form, we can multiply through by 13:

\[ 13\left(\frac{x^2}{52} + \frac{y^2}{13}\right) = 13 \quad \Rightarrow \quad \frac{13x^2}{52} + y^2 = 13 \]

This simplifies to:

\[ \frac{x^2}{4} + y^2 = 13 \]

### Final Answer
The equation of the ellipse is:

\[ \frac{x^2}{4} + \frac{y^2}{13} = 1 \]