If e_{1} and e_{2} represent the eccentricity | vedclass

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(B) The given equations of the curves are $16x^{2}-9y^{2}=144$ and $9x^{2}-16y^{2}=144$. Dividing both by $144$,we get $\frac{x^{2}}{9}-\frac{y^{2}}{1

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(B) The given equations of the curves are $16x^{2}-9y^{2}=144$ and $9x^{2}-16y^{2}=144$. Dividing both by $144$,we get $\frac{x^{2}}{9}-\frac{y^{2}}{16}=1$ and $\frac{x^{2}}{16}-\frac{y^{2}}{9}=1$. For a hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$,the eccentricity $e$ is given by $e = \sqrt{1+\frac{b^{2}}{a^{2}}}$. For the first curve,$a^{2}=9$ and $b^{2}=16$,so $e_{1} = \sqrt{1+\frac{16}{9}} = \sqrt{\frac{25}{9}} = \frac{5}{3}$. For the second curve,$a^{2}=16$ and $b^{2}=9$,so $e_{2} = \sqrt{1+\frac{9}{16}} = \sqrt{\frac{25}{16}} = \frac{5}{4}$. Now,calculating $\frac{1}{e_{1}^{2}}+\frac{1}{e_{2}^{2}} = \frac{1}{(5/3)^{2}} + \frac{1}{(5/4)^{2}} = \frac{9}{25} + \frac{16}{25} = \frac{25}{25} = 1$.

If $e_{1}$ and $e_{2}$ represent the eccentricity of the curves $16x^{2}-9y^{2}=144$ and $9x^{2}-16y^{2}=144$ respectively,then $\frac{1}{e_{1}^{2}}+\frac{1}{e_{2}^{2}}$ is equal to

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