# Equation of chord of a parabola whose midpoint is given

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.

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.