Two fixed are A (a,0) and B(-a,0) If angle A-angle B=theta then the locus of point C on triangle ABC

Dear pallavi

tanA =y/(a-x)

tanB =y/(a+x)

now  tan(A-B) =( tanA -tanB) /(1+tanAtaB)

        tanΘ  = (y/a-x   - y/a+x) /{1+ y²/(a²-x²)}

      tanΘ  = 2xy /{a²-x²+ y²}

    tanΘ (y²-x²) -2xy  +a² tanΘ =0

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