<s>[INST] Equation of parabola which touches a line and coordinate axis 
Equation of parabola which touches $y=x$ line at $(1,1)$ and touches $x$ axis at $(1,0)$

Try: let focus of parabola be $S(p,q)$ and equation of directrix be $y=mx+c$ and a point $P(x,y)$ on parabola.
The definition of parabola $$(x-p)^2+(y-q)^2=\displaystyle \frac{mx-y+c}{\sqrt{1+m^2}}$$
Parabola passes through $(1,0)$ and $(1,1)$
So $$(1-p)^2+q^2=\frac{m-c}{\sqrt{1+m^2}}$$
and $$(1-p)^2+(1-q)^2=\frac{m-1+c}{\sqrt{1+m^2}}$$
Could some help me how to solve it with shorter way. Thanks
 [/INST] When you render the parabola in the form
$(ax^2+bxy+cy^2)+(dx+ey)+f=0$
the quadratic terms make a completed square ($b^2-4ac=0$).  Exploiting this we may use a simplified form, noting the coefficient of $x^2$ must be nonzero since the parabola does not have a horizontal symmetry axis:
$(x+ay)^2+(bx+cy)+d=0$; $x+ay$ and $bx+cy$ must be linearly independent for a true parabola.
Now a tangent point to the $x$ axis at $(1,0)$ means if we put $y=0$ there must be a double root at $x=1$.  So
$x^2+bx+d=0$, $(x-1)^2=x^2-2x+1=0$
Then $b=-2, d=1$.
Now apply the tangency point at $(1,1$).  There must be a double root at $x=1$ (the point of tangency) when $y=x$ (the tangent line) so
$(x+ax)^2+(bx+cx)+d=(1+a)^2x^2+(c-2)x+1=0$
$x^2-2x+1=0$
We must have $c-2=-2$, so $c=0$; and $(1+a)^2=1$ so $a\in\{0,-2\}$.  We then have two candidate equations:
$x^2-2x+1=0$ but this is just the line $x=1$; fails.
$(x-2y)^2-2x+1=0$, this is really a parabola because $x-2y$ and $x$ are linearly independent; good.
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