Eccentricity of ellipse `(x^(2))/(a^(2)) +(y^(2))/(b^(2))=1` if it passes through point (9, 5) and (12, 4) is

Options

A. `sqrt(3//4)`
B. `sqrt(4//5)`
C. `sqrt(5//6)`
D. `sqrt(6//7)`

Answer

To find the eccentricity of the ellipse given by the equation

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

that passes through the points (9, 5) and (12, 4), we will follow these steps:

### Step 1: Substitute the first point (9, 5) into the ellipse equation

Substituting \(x = 9\) and \(y = 5\):

\[ \frac{9^2}{a^2} + \frac{5^2}{b^2} = 1 \]

Calculating the squares:

\[ \frac{81}{a^2} + \frac{25}{b^2} = 1 \quad \text{(Equation 1)} \]

### Step 2: Substitute the second point (12, 4) into the ellipse equation

Substituting \(x = 12\) and \(y = 4\):

\[ \frac{12^2}{a^2} + \frac{4^2}{b^2} = 1 \]

Calculating the squares:

\[ \frac{144}{a^2} + \frac{16}{b^2} = 1 \quad \text{(Equation 2)} \]

### Step 3: Set up the equations for elimination

From Equation 1:

\[ \frac{81}{a^2} + \frac{25}{b^2} = 1 \]

From Equation 2:

\[ \frac{144}{a^2} + \frac{16}{b^2} = 1 \]

### Step 4: Eliminate one variable

We can manipulate these equations to eliminate one of the variables. Let's multiply Equation 1 by \(b^2\) and Equation 2 by \(b^2\):

1. \(81b^2 + 25a^2 = a^2b^2\)
2. \(144b^2 + 16a^2 = a^2b^2\)

Setting them equal to each other:

\[ 81b^2 + 25a^2 = 144b^2 + 16a^2 \]

### Step 5: Rearranging the equation

Rearranging gives:

\[ 25a^2 - 16a^2 = 144b^2 - 81b^2 \]

This simplifies to:

\[ 9a^2 = 63b^2 \]

### Step 6: Solve for the ratio of \(b^2\) to \(a^2\)

Dividing both sides by \(9a^2\):

\[ \frac{b^2}{a^2} = \frac{9}{63} = \frac{1}{7} \]

### Step 7: Find the eccentricity

The eccentricity \(e\) of the ellipse is given by:

\[ e = \sqrt{1 - \frac{b^2}{a^2}} \]

Substituting the ratio we found:

\[ e = \sqrt{1 - \frac{1}{7}} = \sqrt{\frac{6}{7}} \]

### Final Result

Thus, the eccentricity of the ellipse is:

\[ e = \sqrt{\frac{6}{7}} \]