Question

If the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ meets the line $\frac{x}{7}+\frac{y}{2 \sqrt{6}}=1$ on the $x$-axis and the line $\frac{x}{7}-\frac{y}{2 \sqrt{6}}=1$ on the $y$-axis, then the eccentricity of the ellipse is

• A. $\frac{5}{7}$
• B. $\frac{2 \sqrt{6}}{7}$
• C. $\frac{3}{7}$
• D. $\frac{2 \sqrt{5}}{7}$

Answer 1

The Correct Option is A

To find the eccentricity of the ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\), we need to analyze the points where this ellipse intersects the given lines. The lines provided in the problem are:

• \(\frac{x}{7}+\frac{y}{2 \sqrt{6}}=1\), which meets the ellipse on the x-axis.
• \(\frac{x}{7}-\frac{y}{2 \sqrt{6}}=1\), which meets the ellipse on the y-axis.

The given lines represent two intersecting lines which can be simplified to understand their intersections with the axes:

1. Line on x-axis: \(\frac{x}{7} + \frac{y}{2 \sqrt{6}} = 1\) implies \(y = 0\), giving \(x = 7\).
2. Line on y-axis: \(\frac{x}{7} - \frac{y}{2 \sqrt{6}} = 1\) implies \(x = 0\), giving \(y = 2\sqrt{6}\).

Therefore, the ellipse intersects the x-axis at (7,0) and the y-axis at (0,2\(\sqrt{6}\)).

The standard equation of the ellipse is \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\), where \(a\) is the semi-major axis and \(b\) is the semi-minor axis. The x-intercept gives us \(x = a = 7\) and the y-intercept gives us \(y = b = 2\sqrt{6}\).

Now, we calculate the eccentricity \(e\) of the ellipse. The formula for eccentricity \(e\) is given by:

\(e = \sqrt{1 - \frac{b^2}{a^2}}\)

• \(a = 7\) (since the x-intercept is 7)
• \(b = 2\sqrt{6}\) (since the y-intercept is \(2\sqrt{6}\))

Substituting the values of \(a\) and \(b\) into the eccentricity formula, we get:

\(e = \sqrt{1 - \frac{(2\sqrt{6})^2}{7^2}}\)

\(e = \sqrt{1 - \frac{24}{49}}\)

\(e = \sqrt{\frac{49 - 24}{49}}\)

\(e = \sqrt{\frac{25}{49}}\)

\(e = \frac{5}{7}\)

Therefore, the eccentricity of the ellipse is \(\frac{5}{7}\).