The parabola with directrix x+2y−1=0 and focus (1,0) is

The distance $PS$ is given by the distance formula:

$$PS = \sqrt{(x - 1)^2 + (y - 0)^2}$$ $$PS = \sqrt{(x - 1)^2 + y^2}$$

The distance $PL$ from a point $(x_0, y_0)$ to a line $Ax + By + C = 0$ is given by $PL = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$.
For the directrix $x + 2y - 1 = 0$, we have $A=1$, $B=2$, $C=-1$.

$$PL = \frac{|x + 2y - 1|}{\sqrt{1^2 + 2^2}}$$ $$PL = \frac{|x + 2y - 1|}{\sqrt{1 + 4}}$$ $$PL = \frac{|x + 2y - 1|}{\sqrt{5}}$$

According to the definition of a parabola, $PS = PL$.

$$\sqrt{(x - 1)^2 + y^2} = \frac{|x + 2y - 1|}{\sqrt{5}}$$

Squaring both sides to eliminate the square root and absolute value:

$$(x - 1)^2 + y^2 = \frac{(x + 2y - 1)^2}{5}$$

Multiply both sides by 5:

$$5((x - 1)^2 + y^2) = (x + 2y - 1)^2$$

Expand the left side:

$$5(x^2 - 2x + 1 + y^2) = 5x^2 - 10x + 5 + 5y^2$$

Expand the right side using the formula $(a+b+c)^2 = a^2+b^2+c^2+2ab+2ac+2bc$:

$$(x + 2y - 1)^2 = x^2 + (2y)^2 + (-1)^2 + 2(x)(2y) + 2(x)(-1) + 2(2y)(-1)$$ $$= x^2 + 4y^2 + 1 + 4xy - 2x - 4y$$

Now, set the expanded left side equal to the expanded right side:

$$5x^2 - 10x + 5 + 5y^2 = x^2 + 4y^2 + 1 + 4xy - 2x - 4y$$

Move all terms to one side of the equation:

$$5x^2 - x^2 + 5y^2 - 4y^2 - 4xy - 10x + 2x + 4y + 5 - 1 = 0$$

Combine like terms:

$$4x^2 + y^2 - 4xy - 8x + 4y + 4 = 0$$

Rearranging the terms in standard form ($x^2$, $xy$, $y^2$, $x$, $y$, constant):

$$4x^2 - 4xy + y^2 - 8x + 4y + 4 = 0$$