If e_1 and e_2 are respectively the eccentric | vedclass

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(B) The given equations are:$(i) \ 9x^2 - 16y^2 = 144 \Rightarrow \frac{x^2}{16} - \frac{y^2}{9} = 1$$(ii) \ 9x^2 - 16y^2 = -144 \Rightarrow \frac{y^2

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(B) The given equations are:$(i) \ 9x^2 - 16y^2 = 144 \Rightarrow \frac{x^2}{16} - \frac{y^2}{9} = 1$$(ii) \ 9x^2 - 16y^2 = -144 \Rightarrow \frac{y^2}{9} - \frac{x^2}{16} = 1$Equation $(i)$ represents a hyperbola and equation $(ii)$ represents its conjugate hyperbola.For a hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,the eccentricity $e_1$ is given by $e_1^2 = 1 + \frac{b^2}{a^2} = 1 + \frac{9}{16} = \frac{25}{16}$.For the conjugate hyperbola $\frac{y^2}{b^2} - \frac{x^2}{a^2} = 1$,the eccentricity $e_2$ is given by $e_2^2 = 1 + \frac{a^2}{b^2} = 1 + \frac{16}{9} = \frac{25}{9}$.We know that for conjugate hyperbolas,$\frac{1}{e_1^2} + \frac{1}{e_2^2} = 1$.Substituting the values: $\frac{1}{25/16} + \frac{1}{25/9} = \frac{16}{25} + \frac{9}{25} = \frac{25}{25} = 1$.Now,$\frac{1}{e_1^2} + \frac{1}{e_2^2} = \frac{e_1^2 + e_2^2}{e_1^2 e_2^2} = 1$.Therefore,$\frac{e_1^2 e_2^2}{e_1^2 + e_2^2} = 1$.

If $e_1$ and $e_2$ are respectively the eccentricities of the curves $9x^2 - 16y^2 - 144 = 0$ and $9x^2 - 16y^2 + 144 = 0$,then find the value of $\frac{e_1^2 e_2^2}{e_1^2 + e_2^2}$.

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