How to plot the graph of parabola?

Question

I am new in the conics so there are two confusions: 1. How to plot the graph of this parabola equation $y^2-x-2y-1=0$? 2. what would be the equation of parabola when vertex $(3,2)$ and ends of focal chord are $(5,6)$ and $(5, -2)$?

Thanks

Answer 1

Problem $1$:

The equation is: $$x=y^2-2y-1$$ Complete the square: $$x=y^2-2y+1-1-1$$ $$x=(y-1)^2-2$$ The parabola will have the vertex at $(-2, \ 1)$. It will be "sideways" (because the $y$ is squared) and will open to the right (because the coefficient of the $y^2$ term is positive). The graph should look something like:

Problem $2$:

This "focal chord" should be the latus rectum. This means that it is parallel to the directrix. How do I know this? The focal chord is a vertical line segment. I know that the directrix will also be vertical. Anyways, this means that the focus will be at the midpoint of the focal chord. The midpoint is: $$\left(\frac{5+5}{2}, \ \frac{6+(-2)}{2}\right)$$ $$\text{Which is }(5, \ 2)$$ I know that the parabola will be a "sideways" one. We know that the vertex is $(3, \ 2)$. This means that the distance between the focus and the vertex is $2$.

Standard equation of sideways parabola: $$4p(x-h)=(y-k)^2$$ Where $p$ is the distance between the focus and vertex. $$4(2)(x-3)=(y-2)^2$$ $$8(x-3)=(y-2)^2$$ $$8x-24=y^2-4y+4$$ $$\color{green}{y^2-8x-4y+28=0}$$ Hope I helped

Answer 2

Rewrite it: $x = y^2 - 2y - 1 = (y-1)^2 - 2 \to x+2 = (y-1)^2$, and here are the steps to plot the graph:

1. Plot the vertex $V = (-2,1)$.

2. Draw a parabola open to the right having the vertex at $V$.

For the equation of the parabola with the added information of the focal chord being $6-(-2)=8$, then $a = \dfrac{1}{f} = \dfrac{1}{8}$. So: $x-3 = a(y-2)^2 = \dfrac{1}{8}\times (y-2)^2$.

In general, if $V = (a,b)$, and the focal chord length is $f$, then the equation of the parabola is:

$x-a = \dfrac{1}{f}\times (y-b)^2$