Find equation of Parabola when Focus and tangent to the vertex is given. Focus is $(1,1)$ and equation of the tangent is $x+y=1$.
With this information, we have to find equation of parabola and length of latus rectum.
(Solution is given as $x² -2xy+y² -4x -4y+4=0$)
 A: A parabola’s vertex is midway between its focus and directrix. You’re given the tangent line at the vertex, which is parallel to the directrix, so it should be a straightforward matter to find an equation for the directrix: you have its slope/normal and you can find a point on it by reflecting the given focus point $(x_f,y_f)$ in the tangent line. If you put the resulting equation in the form $ax+by+c=0$, then an equation for the parabola is $$(x-x_f)^2+(y-y_f)^2={(ax+by+c)^2\over a^2+b^2}.$$ This equation just expresses the definition of a parabola as the set of points that are equidistant from the focus and directrix. Rearrange this into whatever form is required.  
As for the latus rectum, you don’t even need the equation of the parabola to compute its length. You can get the focal distance from the information given, and the latus rectum length is a fixed multiple of this distance.
A: Got the solution on another thread, thanks to Shashank.
Find slope of the given line (tangent). Say it is m. So the slope of line (axis) joining the point and its mirror image is -1/m. Use slope point form to find equation of the line (axis) and find its interaection with given line (tangent). Finally use the intersection point (vertex) in midpoint formula to get the required point.
Original post
A: HINT:
If the equation of the directrix is $ax+by+c=0$
the equation of the required parabola:  $$(x-1)^2+(y-1)^2=\dfrac{(ax+by+c)^2}{a^2+b^2}$$
Put $y=1-x$ to find a quadratic equation in $x,$ whose roots represent the abscissa of the intersection.
Now for tangency, both roots will be same.
Can you take it from here?
