Midpoint Formula: $$ {\bigg (} \frac{x_1+x_2}{2} , \frac{y_1+y_2}{2} {\bigg )}$$
Distance Formula $$ d = \sqrt{(x_2 - x_1)^2 + (y_2-y_1)^2} $$
Finding a perpendicular bisector:
$$ m_\perp = -\frac{1}{m} $$To find a perpendiculat bisector, use point slope form with the midpoint of the line, and the perpendicular slope $m_\perp$
Up-down parabola: $y = ax^2$
Left-right parabola: $x = ay^2$
If the y is squared, it opens left or right. If the x is squared, it opens up or down.
Vertex form:
$y-k = a(x-h)^2$
$x-h = a(y-k)^2$
All points on the parabola are equidistant from the focal point and the Directrix. The focal point is $p$ distance from the vertex, in the direction that the parabola opens. The directrix runs perpendicular to how the parabola opens, and does not intersect the parabola. It is $p$ distance from the vertex in the opposite direction of the focal point.
$a = \frac{1}{4p} $
Vertex: $(h, k+p)$
Directrix: $y = k-p$
For up-down parabolas:
$ y = \pm\frac{1}{4p}x^2$
$ x = \pm\frac{1}{4p}y^2$
Parabolas with Vertex (h,k)
$$ y-k = \pm\frac{1}{4p}(x-h)^2 $$$$ x-h = \pm\frac{1}{4p}(y-k)^2 $$Definition of a circle:
All points are equidistant from a given point
$(x-h)^2 + (y-k)^2 = r^2$ where the center of the circle is $(h,k)$ and the radius is $r$
$$ \bigg(\frac{x-h}{r_x}\bigg)^2 + \bigg(\frac{y-k}{r_y}\bigg)^2 = 1$$For every point along the ellipse, the sum of the distances from the point to both foci is constant
Where the center is at $(h,k)$. The radius along the $x$ axis is $r_x$ and the radius along the $y$ axis is $r_Y$.
Focal Distance:
$C^2 = big^2-small^2$
where C is the focal distance (the distance from the center to the foci which are along the axis that the ellipse stretches the most along). $big$ and $small$ are the big and small radii of the ellipse
where the center is $(h,k)$. the focal distance is $c^2 = r_x^2 + r_y^2$. The equation of the asymptotes is $(y-k) = \pm\frac{r_y}{r_x}(x-h)$
What could happen?
With a parabola and an ellipse, there can be 1, 2, 3, or 4 solutions