Focus of parabola with two tangents 
A parabola touches x-axis at $(1,0)$ and $y=x$ at $(1,1)$. Find its focus.

My attempt : All I can say is that as angle subtended by this chord at focus is $90^\circ$ as angle between tangents is $45^\circ$. I can find equation of directrix by taking mirror image of focus in tangents and then use the fact that distance from focus is same as from directrix. But, this will get dirty. 
Any help will be appreciated.

I have a different approach to this problem. Using the general conic equation
$$ A x^2 + B x y + C y^2 + D x + E y + F =0 $$
and its derivative (for tangents)
$$ {\rm d} x ( 2 A x + B y + D ) + {\rm d} y ( B x + 2 C y + E) =0 $$
With the parabola constraint $B^2 = 4 A C$, and the following 4 constraints, all coefficients can be found
$$ \begin{align} 
A + D + F & =0 & & \mbox{curve} & (x=1,y=0) \\
A + B + C + D +E + F & = 0 & & \mbox{curve} & (x=1,y=1) \\
2 A + D & =0 & & \mbox{tangent} & (x=1,y=0,{\rm d}x=1,{\rm d}y=0) \\
2 A + 2 B+ 2 C + D + E & =0 & & \mbox{tangent} & (x=1,y=1,{\rm d}x=1,{\rm d}y=1) \\
\end{align} $$
The above is solved for $$C=-B\\D=-2 A\\E=0\\F=A$$
$$ A x^2 -2 A x - B y^2 + B x y +A =0 $$ and $B^2 = 4 A C = -4 A B \} B = -4 A$
$$ A \left( x^2 - 4 x y - 4 y^2 - 2 x + 1 \right) = 0  \\ A = 1 \mbox{ arbitrary}$$
To find the center, the quadratic is at an extrema ( slope is zero ) at
$$ 2 A x_c + B y_c +D =0 \\ B x_c + 2 C y_c +E =0 $$
$$ 
x_c = \frac{2 C D - B E}{B^2 - 4 A C} = \frac{4 A}{4 A +B} \\ 
y_c = \frac{2 A E - B D}{B^2 - 4 A C} = \frac{2 A}{4 A +B} 
$$
with solution for the parabolic constraint $B=0$ yielding
$$ x_c = 1 \\ y_c = \frac{1}{2} $$
