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
Verified by Experts
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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