The parabola has a vertical axis of symmetry. Given two points and the $y$ coordinate of the vertex, how to determine its formula?
For example:
[Math] Formula of parabola from two points and the $y$ coordinate of the vertex
conic sections
Related Solutions
Let's use the following form for the parabola:
$$y=a(x-h)^2+k$$
Distance from vertex $(h,k)$ to your directrix and focus is calculated using the following formula where $p$ is the distance.
$$a=\frac{1}{4p}$$
If $a$ is positive, the equation for your directrix will be as follows:
$$y=k-p$$
Also, the coordinates of the focus will be the following with $+a$:
$$(h,k+p)$$
If $a$ is negative, there are simply a few sign changes. Directrix equation with $-a$:
$$y=k+p$$
Focus with $-a$:
$$(h, k-p)$$
Hopefully this helped.
One uses Pascal's theorem for hexagons inscribed in conics. The hexagons do not need to be convex and embedded, but the order of the points (following cyclicity) is important.
You have two points on the parabola $P_1$ and $P_2$ and the axis $L$ of the parabola. Then the point at infinity $P_{\infty}$ of on $L$ is also on the parabola. Reflect points $P_1$ and $P_2$ with respect to $L$ and obtain the points $P'_1$ and $P'_2$ respectively, which also lie on the parabola.
Big Step 1. Construct the tip of the tip of the parabola $P_0$, i.e. the point where the parabola intersects the parabola's axis $L$, together with the line $b$ through $P_0$ orthogonal to $L$. The line $b$ is the tangent to the parabola at point $P_0$.
Given the five points $P_1, \, P_2, \, P'_1, \, P'_2$ and $P_{\infty}$ and the line $L$ you can construct the sixth $P_0$ using Pascal's theorem for the hexagon $P_{\infty}P_2P'_2P'_1P_1P_0$ (again, the order is importan).
Denote by $P_{\infty}P_2$ the line through $P_2$ parallel to $L$. Construct $Q_1 = P_{\infty}P_2 \cap P_1P_1'$;
Construct $Q_0 = L \cap P_1'P_2'$;
Line $Q_0Q_1$ is Pascal's line for hexagon $P_{\infty}P_2P'_2P'_1P_1P_0$.
Construct $Q_2 = Q_0Q_1 \cap P_2P_2'$;
Then construct point $P_0 = P_1Q_2 \cap L$ which is the sought point (Pascal's theorem). Draw line $b$ through $P_0$ orthogonal to axis $L$.
Big Step 2. Construct that tangent $t_2$ to the parabola at point $P_2$. To do that one can use the degenerate version of Pascal's theorem where the hexagon is $P_0P_0P_1P_2P_2P_2'$ where the line defined by the degenerate edge $P_0P_0$ is tangent line $b$ at $P_0$ and the line defined by the degenerate edge $P_2P_2$ is tangent line $t_2$ at $P_2$.
As already constructed, point $Q_2 = P_0P_1 \cap P_2P_2'$;
Construct $\hat{Q} = P_1P_2 \cap P_0P_2'$
Line $\hat{Q}Q_2 $ is the Pascal line for degenerate hexagon $P_0P_0P_1P_2P_2P_2'$;
Construct $M = \hat{Q}Q_2 \cap b$;
Construct line $t_2 = MP_2$ which is the tangent to the parabola at point $P_2$ (Pascal's theorem, degenerate version).
Concluding Big Step 3.
Draw line $m$ passing through point $M$ and orthogonal to line $t_2$.
Construct the point $F = m \cap L$. This is the focus of the parabola.
Reflect point $F$ with respect to line $b$ and obtain the point $S$ on $L$ such that $SP_0 = FP_0$, i.e. $P_0$ is the midpoint of segment $FS$.
Construct the line $\Gamma$ through $S$ orthogonal to axis $L$. This is the directrix.
Best Answer
Hints: I will first give a straight-forward method, and then give a cleverer method.
Straightforward method:
You know the formula for a parabola is $$y=ax^2+bx+c.$$ The idea now is just to plug in your points and solve the resulting system of equations. The nonvertex points are easy to deal with - they give you the equations
$$6=25a+5b+c,$$ $$8=64a+8b+c.$$
Now you need to deal with the vertex point. Recall the $x$ coordinate of the vertex is $-\frac{b}{2a}$, so we can plug this in to get the final equation we need:
$$10=\frac{b^2}{4a^2}a-\frac{b}{2a}b+c=-\frac{b^2}{4a}+c$$
Cleverer method:
This time we realize that we can write the parabola in a completed square form. That is we can write $$y=a(x+b)^2+c$$ This is helpful because we know that the $y$-coordinate of the vertex corresponds to when the squared term $(x+b)^2=0$ - in other words the place where the parabola reaches an extremum. Hence we directly have $c=10$. Now we can plug in the other points as before and have an easier system to solve: $$6=a(5+b)^2+10,$$ $$8=a(8+b)^2+10$$