Question

Eccentricity of ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ if it passes through point $(9, 5)$ and $(12, 4)$ is:

• A. \( \sqrt{\frac{3}{4}} \)
• B. \( \sqrt{\frac{4}{5}} \)
• C. \( \sqrt{\frac{5}{6}} \)
• D. \( \sqrt{\frac{6}{7}} \)

Hint:

The eccentricity of an ellipse is given by \( e = \frac{c}{a} \), where \( c^2 = a^2 - b^2 \).

Answer 1

The Correct Option is D

Step 1: Equation of the ellipse. The general equation of the ellipse is: \[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \] 
Where: - \(a\) is the semi-major axis. - \(b\) is the semi-minor axis. - The eccentricity \( e \) of the ellipse is given by \( e = \sqrt{1 - \frac{b^2}{a^2}} \). 
Step 2: Substituting the coordinates into the equation. The ellipse passes through the points (9, 5) and (12, 4), so we substitute these into the equation of the ellipse. 
- For point (9, 5): \[ \frac{9^2}{a^2} + \frac{5^2}{b^2} = 1 \] This simplifies to: \[ \frac{81}{a^2} + \frac{25}{b^2} = 1 \quad {(Equation 1)} \] 
- For point (12, 4): \[ \frac{12^2}{a^2} + \frac{4^2}{b^2} = 1 \] This simplifies to: \[ \frac{144}{a^2} + \frac{16}{b^2} = 1 \quad {(Equation 2)} \] 
Step 3: Solving the system of equations. Now, we solve the system of two equations: \[ \frac{81}{a^2} + \frac{25}{b^2} = 1 \quad {(Equation 1)} \] \[ \frac{144}{a^2} + \frac{16}{b^2} = 1 \quad {(Equation 2)} \] Solving these equations gives the values of \(a^2\) and \(b^2\). 
Step 4: Calculating the eccentricity. Once we have the values of \(a^2\) and \(b^2\), the eccentricity is calculated as: \[ e = \sqrt{1 - \frac{b^2}{a^2}} \] 
Step 5: Final Answer. The calculated eccentricity is \( e = \sqrt{\frac{6}{7}} \), which corresponds to Option D.