Since the parabola is parallel to 0x, the curve must be opening to either left or right.
To find the equation using the given points;
Create Equations
Use the equations to calculate \(a, h ~and ~ k\)
\((4-k)^2 =4a(5-h) \)
\(16 - 8k + k^2 = 20a -4ah\) \(eq. ~1\)
\((-2-k)^2 =4a(11-h) \)
\(4 +4k + k^2 = 44a -4ah\) \(eq. ~2\)
\((-4-k)^2 =4a(21-h) \)
\(16+8k + k^2 = 84a -4ah\) \(eq. ~3\)
Remove \(h\) using equations \(1 \) and \(2\)
\begin{align}(16 - 8k + k^2)-(4 +4k + k^2) \\= (20a -4ah)-(44a -4ah)\end{align}
\(12-12k = -24a\) \(eq.~4\)
Remove \(h\) using equations \(2 \) and \(3\)
\begin{align}(4 +4k + k^2) - (16+8k + k^2 ) \\= (44a -4ah) - (84a -4ah)\end{align}
\(-12 - 4k = -40a\) \(eq.~5\)
Find \(k ~ and ~ a\) using equations \(5~ and ~4\)
\(12-12k = -24a)\) \(eq.~4\)
\(36 + 12k = 120a~~\) \(eq.~5\) multiplied by \(-3\)
\(48 = 96a\)
\(a = \frac{1}{2}\)
Solve \(k\)
\(12-12k = -24 \frac{1}{2}\)
\(k=2\)
Use any equations \((1-3)\) to solve \(h\).
\((4-2)^2 = 4(0.5)(5-h)\)
\(h = 3\)
The equation:
\((y-2)^2 = 2(x-3)\) Answer
It can also be written in general form.
\(y^2-4y + 4 = 2x-6\)
\(y^2-4y-2x+10=0\) Answer