Let the foci of a hyperbola H coincide with the foci of the ellipse \[E: \frac{(x-1)^2}{100}+\frac{(y-1)^2}{75}=1\] and the eccentricity of the hyperbola H be the reciprocal of the eccentricity of the ellipse E. If the length of...
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Official solution
E:100(x−1)2+75(y−1)2=1\begin{gathered} {e}_{E} =\sqrt{1-\frac{{b}^{2}}{{a}^{2}}} \\ =\sqrt{1-\frac{75}{100}} \\ {e}_{E}=\frac{1}{2} ∴eH=2 [ as eccentricity of hyperbola is reciprocal of eccentricity of ellipse]
Transverse axis of hyperbola =α
Conjugate axis of hyperbola =β
Also, foci of ellipse (1±ae,1)=(1±(10×21),1)=(1±5,1)=(6,1)(−4,1)
Distance between foci =102ae=10⇒a=25 {e}^{2}=1+\frac{{b}^{2}}{{a}^{2}} \\ 4 =1+\frac{4{b}^{2}}{25} \\ {b}^{2} =\frac{75}{4} \\ b =\frac{\sqrt{75}}{2} \\ \Rightarrow \, \alpha =5 \\ \beta =\sqrt{75} \\ 3{\alpha }^{2}+2{\beta }^{2}=3(5{)}^{2}+2(75)=225 \end{gathered}