Question

if the extremities of a focal chord of the parabola y2=4ax be (at square,2at) and (at square,2at),show that t1t2= -1

Answer 1

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Answer 2

If the extremities of a focal chord of the parabola $y^{2}=4ax$ are $(at_{1}^{2},2at_{1}),(at_{2}^{2},2at_{2})$, then $t_{1}t_{2}=-1$

Proof:

             Equation of parabola is $y^{2}=4ax$

                       Focus = (a,0)

             Equation of chord passing through $(at_{1}^{2},2at_{1}),(at_{2}^{2},2at_{2})$ is given by

              $\frac{y-y_{1}}{y_{2}-y_{1}}=\frac{x-x_{1} }{x_{2}-x_{1}}\\\frac{y-2at_{1}}{2at_{2}-2at_{1}}=\frac{x-at_{1}^{2}}{at_{2}^{2}-at_{1}^{2}}$

           Focal chord passes through focus (a, 0)

                (x, y) = (a, 0)

              $\frac{0-2at_{1} }{2a(t_{2}-t_{1}) }=\frac{a-at_{1}^{2}}{a(t_{2}^{2}-t_{1}^{2})}\\\frac{-2at_{1} }{2a(t_{2}-t_{1})}=\frac{a(1-t_{1}^{2})}{a(t_{2}+t_{1})(t_{2}-t_{1})}\\-t_{1}(t_{2}+t_{1})=1-t_{1}^{2}\\-t_{1}t_{2}-t_{1}^{2} =1-t_{1}^{2}\\-t_{1}t_{2}=1\\t_{1}t_{2}=-1$