Finding the domain of the parabola with focus $(1,2)$ and directrix $2x+y=1$.

Question

Problem. Find the domain of the parabola with focus $(1, 2)$ and directrix $2x+y=1$.

My attempt

Using the distance from a point to a line formula and the point-to-point distance formula, I have gotten this equation: $$\frac{\left|2x+y-1\right|}{\sqrt{5}}=\sqrt{\left(x-1\right)^{2}+\left(y-2\right)^{2}}.$$ Clearly, simplifying this would be a pain and I'm not sure it would help even if I did.

By graphing the parabola on desmos, we find that domain is $\{x|x\ge\frac14\}$. When I try to plug in smaller values such as $\frac15$, I get $$\frac{|\frac25+y-1|}{\sqrt5}=\sqrt{\left(\frac15-1\right)^2+(y-2)^2}.$$ I am pretty sure simplifying and solving would yield no real solutions for $y$, but I want

1. An elegant solution(that does not involve a ton of bashing algebra)
2. A method of finding the domain that does not rely on graphing. Thanks!

Answer 1

The simplest method I see, is actually to simplify $$\frac{\left|2x+y-1\right|}{\sqrt{5}}=\sqrt{\left(x-1\right)^{2}+\left(y-2\right)^{2}}$$

$$\implies5x^2-10x+5y^2-20y+25=4x^2+y^2+1+4xy-4x-2y$$

Rewrite, treating it as a quadratic in $y$

$$4y^2+y(-4x-18)+(x^2-6x+4)=0$$

We wish a value of $x$, for which we have at least one unique solution to $y$ $$\implies \Delta\ge0$$ $$\implies(-4x-18)^2-4\cdot(4)\cdot(x^2-6x+24)\ge0$$ $$\implies4x^2+36x+81-4x^2+24x-96\ge0\implies \boxed{x\ge\frac14}$$

Answer 2

Start from your equation, and square it:$$\frac{\left|2x+y-1\right|}{\sqrt{5}}=\sqrt{\left(x-1\right)^{2}+\left(y-2\right)^{2}}\\(2x+y-1)^2=5(x-1)^2+5(y-2)^2$$ Now move everything to one side and write a quadratic in powers of $y$. $$4x^2+y^2+1+4xy-2y-4x=5x^2-10x+5+5y^2-20y+20\\4y^2-y(18+4x)+x^2-6x+24=0$$ You have only one solution for $y$ when the discriminant is zero: $$(18+4x)^2-16(x^2-6x+24)=0\\324+144x+16x^2-16x^2+96x-384=0\\240x=60\\x=\frac14$$

Answer 3

Here is a purely geometric solution.

It is clear that the minimum $x$ value is given by drawing a vertical tangent to the parabola. By the "focus/directrix" property of the parabola we have $DT=FT$. Using the "reflection" property of the parabola and then opposite angles, we have $$\angle FTC=\angle ETS=\angle DTC\ .$$ Hence $\triangle FTC$ is congruent to $\triangle DTC$. Now write $$\alpha=\angle ABO=\tan^{-1}\frac12\ .$$ Then from congruent triangles and parallel lines, $$\angle FCT=\angle DCT=\alpha\ ;$$ but because $F=(1,2)$, also $$\angle FOB=\tan^{-1}\frac12=\alpha\ .$$ Hence $F,C,O$ are collinear, $\angle COB=\alpha$, and it is now easy to see that $\triangle OCA$ is isosceles and $$OM=\frac12OA=\frac14\ .$$

Answer 4

Here's a more-general geometric argument.

Take the parabola's focus to be $F=(p,q)$, and let the directrix pass through points $A=(a,0)$ and $B=(0,b)$. As indicated in @David's solution the goal here is to find the vertical tangent; say that tangent is at point $T$ with $x$-coordinate $t$.

By the focus-directrix definition and the reflection property of the parabola, foot $F'$ of the perpendicular from $T$ to the directrix is also the reflection of $F$ in the tangent line. Note that since the tangent line is vertical, $\overline{FF'}$ is horizontal.

Writing $G$ for the point where $\overline{FF'}$ meets the $y$-axis, we see that $$2\left(t+|F'G|\right) = |FF'|= p+|F'G| \qquad\to\qquad |F'G|=p-2t$$ Now, we simply observe that $\triangle AOB\sim\triangle F'GB$, so that

$$\frac{a}{b}=\frac{p-2t}{q-b} \qquad\to\qquad t=\frac1{2b}(ab+bp-aq) \tag{$\star$}$$

For the problem at hand, $p=1$, $q=2$, $a=1/2$, $b=1$, so $t=1/4$. $\square$