Hard geometry question
2 rod of length and slide on the plane such that point of intersection of lines with and axis are concyclic. The locus of center of the circle is a certain conic. Find its latus rectum.
You are given .
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1 solution
Write equation of both line in intercept point I.e x/a+y/b=1 Apply family L1L2+kxy=0(l1 is equation of line) Then compare coeffient of x^2, y^2 (to have circle coefficient should be same)to get the sum of the both angles made by line to be pi/2 or 90 degrees. Find the locus by taking midpoint of both intersection point of lines with x, y axis to be the center. To eliminate thita first add both eq., then subtract then square and add both to get equation of locus.(put the diff angle as 90-thita) Is form (x+y)^2/(a+b)+(x-y)^2 (a-b)=2 Check whether h^2 <ab to get ellipse To get max and min Coordinate put sin thita to be 1 and -1 in there respective Coordinate ex sin thita ans cos thita are negative in 3rd quadrand. To find latusrectum apply 2*(semi minor axis)^2/(semi major) to get 2×(a-b)^2/(a+b)