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How to prove $$ T = S1 $$ $$ i.e \qquad yy_1 - 2a(x+x_1) = y_1^2 - 4ax_1=0$$ as the equation of chord for a parabola y$^2$ = 4ax whose midpoint (x$_1,y_1$) is given.

$$$$ I couldn't understand how the equation of chord, can be the same as the equation of tangent at $ (x_1,y_1$) i.e $yy_1 - 2a(x+x_1)=0$. Again since there is a tangent at (x$_1,y_1$) that mean we have a parabola inside. If it is so, how we have same focus (a,0) for both the parabola.

Thanks.

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You made a typo here. It should be $$ yy_1-2a(x+x_1)=y_1^2-4ax_1 $$ without the $=0$ at the end. The point $(x_1,y_1)$ does not lie on the parabola $y^2=4ax$. Instead, it is tangent to a shifted parabola $y^2-4ax=y_1^2-4ax_1$ in order to be the midpoint of a chord. The foci of this shifted parabola is $(a+x_1-y^2/(4a),0)$.

Let's get back to proving the equation of chord. Suppose $(as^2,2as)$ and $(at^2, 2at)$ are two points on the parabola $y^2=4ax$. The mid-point of the chord is $$ \left(a\frac{s^2+t^2}{2},a(s+t)\right)=:(x_1,y_1) $$ and the equation of chord is $$ \frac{y-y_1}{x-x_1}=\frac{2a(t-s)}{a(t^2-s^2)}=\frac{2a}{a(t+s)}=\frac{2a}{y_1}. $$ So $$ yy_1-y_1^2=2a(x-x_1), $$ or equivalently $$ yy_1-2a(x+x_1)=y_1^2-4ax_1, $$ as claimed.

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