The line y=x intersects the hyperbola frac{x^ | vedclass

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(D) Given the equations of the hyperbola and the line are $\frac{x^{2}}{9}-\frac{y^{2}}{25}=1$ and $y=x$ respectively.To find the intersection points,

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$\frac{2 \sqrt{2}}{3}$

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(D) Given the equations of the hyperbola and the line are $\frac{x^{2}}{9}-\frac{y^{2}}{25}=1$ and $y=x$ respectively.To find the intersection points, substitute $y=x$ into the hyperbola equation:$\frac{x^{2}}{9}-\frac{x^{2}}{25}=1$$\frac{25x^{2}-9x^{2}}{225}=1$ $\Rightarrow 16x^{2}=225$ $\Rightarrow x^{2}=\frac{225}{16}$$x=\pm \frac{15}{4}$. Since $y=x$, the intersection points are $P\left(\frac{15}{4}, \frac{15}{4}\right)$ and $Q\left(-\frac{15}{4}, -\frac{15}{4}\right)$.The length of the segment $PQ$ (major axis) is the distance between $P$ and $Q$:$2a = \sqrt{(\frac{15}{4} - (-\frac{15}{4}))^{2} + (\frac{15}{4} - (-\frac{15}{4}))^{2}} = \sqrt{(\frac{30}{4})^{2} + (\frac{30}{4})^{2}} = \sqrt{2 \cdot (\frac{15}{2})^{2}} = \frac{15}{\sqrt{2}}$.Thus, $a = \frac{15}{2\sqrt{2}}$.The length of the minor axis is $2b = \frac{5}{\sqrt{2}}$, so $b = \frac{5}{2\sqrt{2}}$.The eccentricity $e$ of the ellipse is given by $e = \sqrt{1 - \frac{b^{2}}{a^{2}}}$.$e = \sqrt{1 - \frac{(5/2\sqrt{2})^{2}}{(15/2\sqrt{2})^{2}}} = \sqrt{1 - (\frac{5}{15})^{2}} = \sqrt{1 - (\frac{1}{3})^{2}} = \sqrt{1 - \frac{1}{9}} = \sqrt{\frac{8}{9}} = \frac{2\sqrt{2}}{3}$.

The line $y=x$ intersects the hyperbola $\frac{x^{2}}{9}-\frac{y^{2}}{25}=1$ at the points $P$ and $Q$. The eccentricity of the ellipse with $PQ$ as the major axis and a minor axis of length $\frac{5}{\sqrt{2}}$ is

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