1.

If the normal at theta on the hyperola x^(2)/a^(2)-y^(2)/b^(2)=1 meets the transverse axis at G, prove that AG, A'G=a^(2) (e^(4) sec^(2) theta-1). Where A and A' are the vertices of the hyperbola.

Answer»


ANSWER :`(-a + AE^(2) sec theta) (a + ae^(2) sec theta) = a^(2) (e^(4) sec^(2) theta -1)`


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