If the ellipse frac{x^{2}}{a^{2}}+frac{y^{2}} | vedclass

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(A) 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. Setting $y=0$ in the l

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$\frac{5}{7}$

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(A) 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. Setting $y=0$ in the line equation,we get $x=7$. Since the ellipse passes through $(7, 0)$,we have $a=7$.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 $y$-axis. Setting $x=0$ in the line equation,we get $y=-2\sqrt{6}$. Since the ellipse passes through $(0, -2\sqrt{6})$,we have $b^{2}=(-2\sqrt{6})^{2}=24$,so $b=2\sqrt{6}$.The eccentricity $e$ is given by $e^{2}=1-\frac{b^{2}}{a^{2}}$.Substituting the values,$e^{2}=1-\frac{24}{49} = \frac{25}{49}$.Therefore,$e=\frac{5}{7}$.

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

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