The eccentricity of an ellipse E with centre at the origin is 23 and its directrices are x=±346. Let H be a hyperbola whose eccentricity is equal to the length of the semi-major axis of E, and whose length of latus rectum is equal to the length of the minor axis of E. Then the distance between the foci of H is:
Solutions · 1
Official solution
For E (major axis along x): directrix x=eA=346 with e=23:
\[A=\frac{\sqrt{3}}{2}\cdot\frac{4\sqrt{6}}{3}=\frac{4\sqrt{18}}{6}=2\sqrt{2} \;\Rightarrow\; A^2=8.\]
\[B^2=A^2(1-e^2)=8\left(\frac14\right)=2.\]
For H: eH=A=22, latus rectum =2B=22.
\[e_H^2=1+\frac{b^2}{a^2}=8 \;\Rightarrow\; \frac{b^2}{a^2}=7 \;\Rightarrow\; b^2=7a^2.\]
\[\frac{2b^2}{a}=2\sqrt{2} \;\Rightarrow\; b^2=a\sqrt{2} \;\Rightarrow\; 7a^2=a\sqrt{2} \;\Rightarrow\; a=\frac{\sqrt{2}}{7}.\]
\[a^2=\frac{2}{49},\quad b^2=7a^2=\frac27,\quad c^2=a^2+b^2=\frac{2}{49}+\frac{14}{49}=\frac{16}{49}\;\Rightarrow\;c=\frac47.\]
\[\text{Distance between foci}=2c=\frac87.\]