The locus of a point whose chord of contact w.r.t. the hyperbola x^2 a^2 - y^2 b^2 =1 touches the circle described on the straight line joining the foci of the hyperbola x^2 a^2 - y^2 b^2 =1 as diameter is

Step-by-step solution

Circle on the join of foci (a e, 0) and (-a e, 0) diameter is: aligned & (x-a e)(x+a e)+(y-0)(y-0)=0 & x^2+y^2=a^2 e^2=a^2+b^2 WC...(1) & [a^2 e^2=a^2+b^2 ] aligned Let chord of contact of P (x₁, y₁ ) touch the circle (1) Equation of chord of contact P is [T=0] aligned & x x₁ a^2 - y y₁ b^2 =1 (2) & b^2 x₁ x-a^2 y₁ y-a^2 b^2=0 & a^3 b^2 (b^4 x₁^2+a^4 y₁^2 ) = (a^2+b^2 ) aligned Hence locus of P (x₁, y₁ ) is (b^4 x^2+a^4 y^2 ) (a^2+b^2 )=a^4 b^4 .