find the equation of parabola whose focus is (1,1) and tangent at the vertex is x + y = 1

Let S be the focus and A be the vertex of the parabola. Let K be the point of intersection of the axis and directrix.

Since axis is a line passing through S(1,1) and perpendicular to x+y=1 so, let the equation of the axis be x-y+λ=0.

This will pass through (1,1), if 1-1+λ=0 ⇒λ=0

So the equation of the axis is x-y=0

The vertex A is the point of intersection of x-y=0 and x+y=1.

Solving these two equations, we get x=1/2 , y=1/2

so, the coordinates of the vertex A are (1/2,1/2).

Let  $\left({\mathrm{x}}_1, {\mathrm{y}}_1\right)$ be the coordinates of K. then,

$$\begin{gathered} \frac{{\mathrm{x}}_1+1}{2} = \frac{1}{2} \Rightarrow {\mathrm{x}}_1 = 0 \\ \frac{{\mathrm{y}}_1+1}{2} = \frac{1}{2} \Rightarrow {\mathrm{y}}_1 = 0 \end{gathered}$$

So, the co-ordinates of K are (0,0).

Since directrix is a line passing through K(0,0) and parallel to x+y=1, 

therefore, equation of the directrix is y-0=-1(x-0), i.e. x+y=0

Let P(x,y) be any point on the parabola. then,

distance of P from the focus S = distance of P from the directrix

$$\begin{gathered} \Rightarrow \sqrt{{\left(\mathrm{x}-1\right)}^2 + {\left(\mathrm{y}-1\right)}^2} = \frac{\left|\mathrm{x}+\mathrm{y} - 0\right|}{\sqrt{1^2 + 1^2}} \\ \Rightarrow {\mathrm{x}}^2+1-2\mathrm{x}+{\mathrm{y}}^2+1-2\mathrm{y} = \frac{{\left(\mathrm{x}+\mathrm{y}\right)}^2}{2} \\ \Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2-4\mathrm{x}-4\mathrm{y}-2\mathrm{xy}+4 = 0 \end{gathered}$$