If PN is the perpendicular from a point on a rectangular hyperbola to its asymptotes, the locus, the midpoint of PN is a/an

Dear Zahid

you have not maintion where is point P and N ,

ia m solving this question asuming P is on the parabola and N is on the asymptotes

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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