Find the equation of the parabola with Foci(1,2) and directrix $y=5x-3$. Write your answer in the form $Ax^2+Bxy+Cy^2=D$. Sketch the graph of the parabola.
My work so far:
the distance between a point on the parabola $(x,y)$ and $(1,2)$ is
$$\sqrt{(x-1)^2+(y-2)^2}$$
The distance between a point on the parabola $(x,y)$ and the line $y=5x-3$ is given by the formula $\frac{|Ax+By+C|}{\sqrt{A^2+B^2}}$, so in this case it would be $$\frac{|-5x+y+3|}{\sqrt{26}}$$
Then I set the distances equal to each other:
$\sqrt{(x-1)^2+(y-2)^2} = \frac{|-5x+y+3|}{\sqrt{26}}$
Squaring both sides:
$(x-1)^2+(y-2)^2 = \frac{(-5x+y+3)^2}{26}$
$26[(x-1)^2+(y-2)^2]=25x^2-10xy-30x+y^2+6y+9$
$26x^2-52x+26y^2-104y+130=25x^2-10xy-30x+y^2+6y+9$
$x^2+10xy-22x+25y^2-110y+121=0$
So that is my answer for the equation of the parabola. My question now is: how do I graph this? I know that the foci and directrix and both the same distance to the vertex, but I do not know how to graph this parabola. Furthermore, I tried graphing the above equation in Desmos, and I just get a line. Where did I go wrong and how do I graph this?
As @RamanujanXV has said in the comments, the parabola is only defined for focus not on the directrix, but the focus $(1, 2)$ is on the directrix $y=5x-3$. So if you try to apply the definition of a parabola(distance to focus is equal to distance to directrix), you will find a straight line perpendicular to the directrix, and passing the point $(1, 2)$, namely $x+5y=11$. The quadratic equation you found is exactly the square of this; $(x+5y-11)^2=0$.