Find the equation of the ellipse of major axis is along x - axis, centre is at origin and it passes though the point (4,3) and (6,2)

Answer

To find the equation of the ellipse with the given conditions, we will follow these steps:

### Step 1: Write the standard form of the ellipse
Since the major axis is along the x-axis and the center is at the origin, the standard form of the ellipse is given by:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]

### Step 2: Substitute the points into the equation
We know that the ellipse passes through the points (4, 3) and (6, 2). We will substitute these points into the equation to create two equations.

For the point (4, 3):
\[ \frac{4^2}{a^2} + \frac{3^2}{b^2} = 1 \]
This simplifies to:
\[ \frac{16}{a^2} + \frac{9}{b^2} = 1 \quad \text{(Equation 1)} \]

For the point (6, 2):
\[ \frac{6^2}{a^2} + \frac{2^2}{b^2} = 1 \]
This simplifies to:
\[ \frac{36}{a^2} + \frac{4}{b^2} = 1 \quad \text{(Equation 2)} \]

### Step 3: Solve the equations simultaneously
Now we have two equations:
1. \(\frac{16}{a^2} + \frac{9}{b^2} = 1\)
2. \(\frac{36}{a^2} + \frac{4}{b^2} = 1\)

We can manipulate these equations to eliminate one variable. Let's multiply Equation 1 by \(b^2\) and Equation 2 by \(b^2\) to get rid of the denominators:
1. \(16b^2 + 9a^2 = a^2b^2\)
2. \(36b^2 + 4a^2 = a^2b^2\)

Now we can subtract these two equations:
\[ (36b^2 + 4a^2) - (16b^2 + 9a^2) = 0 \]
This simplifies to:
\[ 20b^2 - 5a^2 = 0 \]
or
\[ 5a^2 = 20b^2 \implies a^2 = 4b^2 \]

### Step 4: Substitute back to find \(b^2\)
Now substitute \(a^2 = 4b^2\) into Equation 1:
\[ \frac{16}{4b^2} + \frac{9}{b^2} = 1 \]
This simplifies to:
\[ \frac{4}{b^2} + \frac{9}{b^2} = 1 \implies \frac{13}{b^2} = 1 \implies b^2 = 13 \]

### Step 5: Find \(a^2\)
Now, substitute \(b^2\) back to find \(a^2\):
\[ a^2 = 4b^2 = 4 \times 13 = 52 \]

### Step 6: Write the final equation of the ellipse
Now that we have \(a^2\) and \(b^2\), we can write the equation of the ellipse:
\[ \frac{x^2}{52} + \frac{y^2}{13} = 1 \]

### Final Answer:
The equation of the ellipse is:
\[ \frac{x^2}{52} + \frac{y^2}{13} = 1 \]