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
 A: 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.
A: From the two tangents you can find the parabola’s axis direction. It’s the diagonal of the paralellogram formed by the tangents and their intersection point. These two tangents intersect at the origin, so the axis direction is $\mathbf v = (1,1)+(1,0)=(2,1)$.  
You can now use the reflective property of the parabola to find its focus. The direction vector $\mathbf v$ is reflected to $(-2,1)$ at the point $(1,0)$ and to $(1,2)$ at the point $(1,1)$, so the parabola’s focus is the intersection of the lines $(1,0)+\lambda(-2,1)$ and $(1,1)+\mu(1,2)$, which can be found via cross products of homogeneous coordinates: $$((1,0,1)\times(-2,1,0))\times((1,1,1)\times(1,2,0))=(-3,-1,-5),$$ or $F=\left(\frac35,\frac15\right)$ in inhomogeneous Cartesian coordinates.  
The feet of perpendiculars dropped from the focus to tangent lines lie on the tangent to the parabola’s vertex. The point $\left(\frac35,0\right)$ on this line is easily found by inspection. Knowing also that this line is perpendicular to the parabola’s axis, we can form its point-normal equation: $(2,1)\cdot(x,y)-(2,1)\cdot\left(\frac35,0\right)=0$ or $$2x+y-\frac65=0.$$ This line is halfway between the focus and directrix, so we offset the left-hand side by $(2,1)\cdot F-\frac65 = \frac15$ resulting in the equation $$2x+y=1$$ for the directrix. You now have the necessary information to use the form of the equation of a parabola in your question.  

Alternatively, once you have the axis direction, you know that the equation will be of the form $$(x-2y)^2+ax+by+c=0.\tag{*}$$ The normals to this curve can be found by differentiation: $(a+2x-4y,b-4x+8y)$. If the tangent at a point has direction $(\lambda,\mu)$ the condition that the normal is orthogonal to the tangent can be expressed as $$(\lambda,\mu)\cdot(a+2x-4y,b-4x+8y)=0.$$ Plugging in the known points and tangents will give you a system of linear equations in $a$ and $b$. Once you’ve solved for those, you can plug one of the points into equation (*) to find $c$.
A: I am assuming "touches" means "is tangent to".
Note that the general equation for a conic section in the plane is
$$
ax^2 + bxy + cy^2 + dx+ey+f = 0
$$
First off, since we have a parabola, we must have
$$
\boxed{b^2 - 4ac = 0}
$$
(for otherwise it would be an ellipse or a hyperbola). Next, since it passes through the points $(1,1)$ and $(1,0)$, we can insert those coordinates into the general equation and get
$$
\boxed{a+b+c+d+e+f = 0}\\
\boxed{a + d + f = 0}
$$
Finally, we can use the tangents we know. This is easiest done through implicit differentiation: For any small non-vertical part (in practice any of the two non-vertical parts) of the parabola, we can see $y$ as a function of $x$, and that function will still fulfill the equation above. That means we can differentiate with respect to $x$, which yields
$$
2ax + by + bxy' + 2cyy' + d + ey' = 0
$$
Into this we can insert that at the point $(1, 1)$ we have $y' = 1$, and at the point $(1, 0)$, we have $y' = 0$ to get
$$
\boxed{2a+2b+2c+d+e = 0}\\
\boxed{2a + d = 0}
$$
We now have five equations for our six unknowns (the boxed equations). Where is the last equation? There isn't one. So there isn't a unique solution to our equations. This reflects the fact that the equation of an algebraic set (such as a parabola, a circle, or anything else) is homogenous with respect to the coefficients. In other words, we can freely multiply all coefficients simultaneously by any non-zero number, and the new equation still describes the same set.
So as you solve these equations (which shouldn't take too long, despite what you might say in the comments), you will finally get all coefficients defined as some fixed multiple of a single coefficient. In other words, you will get something like the following (these are not the actual solutions)
$$
b = 2a\\
c = -2a\\
d = 0\\
e = 2.5a\\
f = a
$$
(of course, that one quadratic equation will probably make things a bit more interesting, but not much). Once you're at this point, just pick a non-zero number for $a$ (if that's the unknown you choose to express everything with respect to) and find the rest of the coefficients, and you will have your parabola equation.
