A circle touches the parabola y^2=4x at (1,2) | vedclass

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(C) The equation of the parabola is $y^2=4x$. The directrix of this parabola is $x=-1$.The equation of the tangent to the parabola at point $B(1,2)$ i

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$2\sqrt{2}$

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(C) The equation of the parabola is $y^2=4x$. The directrix of this parabola is $x=-1$.The equation of the tangent to the parabola at point $B(1,2)$ is given by $y y_1 = 2a(x+x_1)$,where $a=1$. Substituting $(1,2)$,we get $2y = 2(x+1)$,which simplifies to $y=x+1$.The tangent $y=x+1$ intersects the directrix $x=-1$ at point $A$. Substituting $x=-1$ into the tangent equation,we get $y=-1+1=0$. So,$A$ is $(-1,0)$.Let the circle touch the directrix at point $C(-1, k)$. Since $AB$ and $AC$ are tangents to the circle from the same external point $A$,their lengths must be equal,i.e.,$AB=AC$.The length $AB = \sqrt{(1 - (-1))^2 + (2-0)^2} = \sqrt{2^2 + 2^2} = \sqrt{4+4} = \sqrt{8} = 2\sqrt{2}$.The length $AC = \sqrt{(-1 - (-1))^2 + (k-0)^2} = \sqrt{0^2 + k^2} = |k|$.Equating the lengths,$|k| = 2\sqrt{2}$. Since the point $C$ is above the $x$-axis in the diagram,$k = 2\sqrt{2}$.

$A$ circle touches the parabola $y^2=4x$ at $(1,2)$ and also touches its directrix. The $y$-coordinate of the point of contact of the circle and the directrix is

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