# Thread: Definine a Formula as a Parabola

1. ## Definine a Formula as a Parabola

Hello,

How would I go about showing that the equation (x-a)^2 = x^2 + y^2 defines a parabola? If I put the equation in terms of y, I get y = sqrt(a^2 - 2xa), which is not a parabola. I've found the derivative to be y' = -a/y, but I don't think that proves anything. If anyone has any ideas, I'd really appreciate them! Thanks.

PS 'a' is a constant.

Hello,

How would I go about showing that the equation (x-a)^2 = x^2 + y^2 defines a parabola? If I put the equation in terms of y, I get y = sqrt(a^2 - 2xa), which is not a parabola. I've found the derivative to be y' = -a/y, but I don't think that proves anything. If anyone has any ideas, I'd really appreciate them! Thanks.

PS 'a' is a constant.
$\displaystyle x^2-2ax+a^2=x^2+y^2\Rightarrow\ 2ax=-y^2+a^2\Rightarrow\ x=\displaystyle-\frac{1}{2a}y^2+\frac{a}{2}$

which is a parabola.

Hello,

How would I go about showing that the equation (x-a)^2 = x^2 + y^2 defines a parabola? If I put the equation in terms of y, I get y = sqrt(a^2 - 2xa), which is not a parabola. I've found the derivative to be y' = -a/y, but I don't think that proves anything. If anyone has any ideas, I'd really appreciate them! Thanks.

PS 'a' is a constant.
$\displaystyle x^2 - 2ax + a^2 = x^2 + y^2$

$\displaystyle 2ax = a^2-y^2$

$\displaystyle x = \frac{a}{2} - \frac{1}{2a} y^2$

parabola, line of symmetry is the x-axis ... opens left from vertex $\displaystyle \left(\frac{a}{2} , 0 \right)$

graph shows parabola for $\displaystyle a = 2$

4. y = sqrt(a^2 - 2xa)
Actually, the equation $\displaystyle y = \hspace{2mm}$\pm$\sqrt{a^2 - 2xa}$ is a parabola. You probably don't recognize it as such because it opens to the left, rather than up or down. You can show that it's a parabola by comparing it to the conic section equation: