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Show that the ellipse 7x^{2}+16y^{2}=112\, and the hyperbola {\frac  {x^{2}}{144}}-{\frac  {y^{2}}{81}}={\frac  {1}{25}}\, have the same foci.

From the ellipse a^{2}=16,b^{2}=7\,

7=16(1-e^{2}),e^{2}={\frac  {9}{16}},e={\frac  {3}{4}}\,

Foci of the ellipse is (\pm ae,0)=(\pm 3,0)\,. let this be 1.

From the given hyperbola a^{2}={\frac  {144}{25}},b^{2}={\frac  {81}{25}}\,

{\frac  {81}{144}}+1=e^{2},e^{2}={\frac  {225}{144}},e={\frac  {5}{4}}\,

Foci are (\pm ({\frac  {12}{5}})({\frac  {5}{4}}),0)=(\pm 3,0)\,. Let this equation be 2.

Since 1 and 2 are same, the given ellipse and hyperbola have the same foci.

Main Page:Geometry:Hyperbola