If `(asectheta,btantheta)` and `(asecphi; btanphi)` are the ends of the focal chord of `x^2/a^2-y^2/b^2=1` then prove that `tan(x/a)tan(phi/2)=(1-e)/(1+e)`
Text Solution
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The equation of the chord joining `(a sec theta, b tan theta)` and `(a sec phi, b tan phi)` is `(x)/(a)cos((theta-phi)/(2))-(y)/(b)sin((theta+phi)/(2))=cos((theta+phi)/(2))` This passes through (ae, 0). Therefore, `ecos((theta-phi)/(2))=cos((theta+phi)/(2))` `"or "e=(cos((theta+phi)/(2)))/(cos((theta-phi)/(2)))` `"or "(e-1)/(e+1)=(cos((theta+phi)/(2))-cos((theta-phi)/(2)))/(cos((theta+phi)/(2))+cos((theta+phi)/(2)))` `"or "(e-1)/(e+1)=-tan.(theta)/(2)tan.(phi)/(2)` `"or "tan.(theta)/(2)tan.(phi)/(2)=(1-e)/(1+e)`
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