Badiuddin askIITians.ismu Expert
Last Activity: 14 Years ago
Dear Zahid
let equation of hyperbola is x2 -y2 =a2
and equation os aymptote is : y=x (slope m=1)
let N(x1,x1) on the asymptote ,and a point P(asecθ ,atanθ) on the hyperbola and(h,k) is the mid point of PN
h=(x1+asecθ)/2 or asecθ =2h-x1 ...............1
k = (x1+atanθ)/2 or atanθ =2k-x1 ................2
since PN is perpendiculat to aymptote so slope of PN =-1
(k-x1)/(h-x1)=-1
x1 =(h+k)/2
so put this value in equation 1 and 2
asecθ=(3h-k)/2
atanθ =(3k-h)/2
now square and subtract
a2(sec2θ -tan2θ) = {(3h-k)/2}2 - {(3k-h)/2}2
a2 = {(3h-k)/2}2 - {(3k-h)/2}2
4a2 =8(h2-k2)
h2-k2=a2/2
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