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If the normal at theta on the hyperola x...

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.

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The correct Answer is:
`(-a + ae^(2) sec theta) (a + ae^(2) sec theta) = a^(2) (e^(4) sec^(2) theta -1)`
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