Question

# If eccentricities of two hyperbolas x2a2−y2b2=1 and x2a2−y2b2=−1 are e1 and e2 respectively then prove that 1e21+1e22=1.

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Solution

## The eccentricity e1 of the given hyperbola is obtained fromb2=a2(e21−1) … (1)The eccentricity e2 of the conjugate hyperbola is given bya2=b2(e22−1) … (2)Multiply (1) and (2), we get,1=(e21−1)(e22−1)⟹0=e21e22−e21−e22⟹e−21+e−22=1Hence proved.

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