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If the normal at P(theta) on the hyperbo...

If the normal at `P(theta)` on the hyperbola `(x^2)/(a^2)-(y^2)/(2a^2)=1` meets the transvers axis at `G ,` then prove that `A GdotA^(prime)G=a^2(e^4sec^2theta-1)` , where `Aa n dA '` are the vertices of the hyperbola.

Text Solution

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The equation of the normal at `P(a sec theta, b tan theta) ` to the given hyperbola is `ax cos theta+by cos theta=(a^(2)+b^(2))`
This meets the transverse axis, i.e., the x-axis at G.
So, the coordinates of G are `({a^(2)+b^(2))//a}sec theta,0).`
The corrdinates of the vertices A and A' are (a,0) and `(-a, 0)`, respectively.
`therefore" "AG*A'G=(-a+(a^(2)+b^(2))/(a)sectheta)(a+(a^(2)+b^(2))/(a)sectheta)`
`=(-a+ae^(2)sectheta)(a+ae^(2)sectheta)`
`=a^(2)(e^(4)sec^(2)theta-1)`
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