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.
 A: (Sketch.)  Let $P$ and $Q$ be two points on a parabola, and let $R$ be the point where the respective tangents to the parabola intersect.  Let $X$ the midpoint of $PQ$.  Then $RX$ is parallel to the axis of symmetry of the parabola (proved below).  Draw lines through $P$ and $Q$ parallel to $RX$, and reflect these lines in the respective tangents; the focus $F$ is the intersection of the reflected lines.

To show that $RX$ is parallel to the axis of symmetry: Drop perpendiculars from $P,Q,R$ to the directrix, meeting it at $P',Q',R'$ respectively.  As you alluded to, the tangent at $P$ is the perpendicular bisector of the segment $FP'$, and likewise for $Q$ and $FQ'$.  So, in $\triangle FP'Q'$, two of the perpendicular bisectors pass through $R$; therefore the third does as well.  Since $RR'$ is a line through $R$ and perpendicular to $P'Q'$, it must be the perpendicular bisector, that is, $R'$ is the midpoint of $P'Q'$.  By parallels, (the extension of) $RR'$ bisects $PQ$, that is, $RR'$ passes through $X$.  So $RX$ is perpendicular to the directrix, as claimed.

Edit: Just for reference, here's what this looks like analytically: The direction of $RX$ is $(2,1)$; reflecting in $RP$ just means exchanging $x$ and $y$ coordinates, so the direction of $PF$ is $(1,2)$, and the line through $P$ in that direction is $2x-y=1$.  Reflecting in $RQ$ means negating the $y$-coordinate, so the direction of $QF$ is $(2,-1)$, and the line through $Q$ in that direction is $x+2y=1$.  The intersection of these lines is $(\frac35,\frac15)$.
A: 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
$$ x_c = 1 \\ y_c = \frac{1}{2} $$
