Question

p is the length of the perpendicular drawn from the origin upon a straight line then the locus of mid point of the portion of the line intercepted between the coordinate axes is. Prove that 1/x^2+1/p^2=4/p^2

Answer 1

Let the equation of AB be :    x/a + y/b = 1A = (a,0) and  B = (0, b)     Origin O(0,0)Length of perpendicular OC from O onto AB is p.Then   p^2 = 1 / [1/a^2 + 1/b^2]=> 1/a^2 + 1/b^2 = 1/p^2      --- (1)Coordinates of the mid point P of AB:  x = a/2     y = b/2So    a = 2x  and  b = 2 ySubstitute for a and b in eq (1) to get:   1/x^2 +  1/y^2 = 4/p^2