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Q.If the eccentricity of a hyperbola is 54\dfrac{5}{4}, then find the eccentricity of its conjugate hyperbola.

Yanam BieapBIEAP Intermediate Board 2025Subjective· 2mImportance★★★★★
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If e1,e2e_1,e_2 are the eccentricities of a hyperbola and its conjugate hyperbola, then 1e12+1e22=1\dfrac{1}{e_1^2}+\dfrac{1}{e_2^2}=1.

Given e1=54e_1=\dfrac{5}{4}:

1(5/4)2+1e22=1  ⟹  1625+1e22=1  ⟹  1e22=925.\frac{1}{(5/4)^2}+\frac{1}{e_2^2}=1 \implies \frac{16}{25}+\frac{1}{e_2^2}=1 \implies \frac{1}{e_2^2}=\frac{9}{25}.

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