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If an ellipse with foci at $(3,3)$ and $(-4,4)$ is passing through the origin, then the eccentricity of that ellipse is
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The correct answer is:
$\frac {5}{7}$
$P S_1+P S_2=2 a$
$\begin{array}{ll}\Rightarrow & \sqrt{(3-0)^2+(3-0)^2}+\sqrt{(4-0)^2+(-4-0)^2}=2 a \\ \Rightarrow & 3 \sqrt{2}+4 \sqrt{2}=2 a \\ \Rightarrow & \frac{7 \sqrt{2}}{2}=a \Rightarrow a=\frac{7}{\sqrt{2}}... (i) \\ & \quad 2 a e=S_1 S_2 \\ \Rightarrow & 2 \cdot \frac{7}{\sqrt{2}} \cdot e=\sqrt{(-4-3)^2+(4-3)^2}=\sqrt{49+1} \\ \Rightarrow & e=\frac{\sqrt{50} \cdot \sqrt{2}}{7 \cdot 2} \Rightarrow e=\frac{10}{14}=\frac{5}{7}\end{array}$
$\begin{array}{ll}\Rightarrow & \sqrt{(3-0)^2+(3-0)^2}+\sqrt{(4-0)^2+(-4-0)^2}=2 a \\ \Rightarrow & 3 \sqrt{2}+4 \sqrt{2}=2 a \\ \Rightarrow & \frac{7 \sqrt{2}}{2}=a \Rightarrow a=\frac{7}{\sqrt{2}}... (i) \\ & \quad 2 a e=S_1 S_2 \\ \Rightarrow & 2 \cdot \frac{7}{\sqrt{2}} \cdot e=\sqrt{(-4-3)^2+(4-3)^2}=\sqrt{49+1} \\ \Rightarrow & e=\frac{\sqrt{50} \cdot \sqrt{2}}{7 \cdot 2} \Rightarrow e=\frac{10}{14}=\frac{5}{7}\end{array}$
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