0

Suppose I have a parabolic equation: $y=-\frac{1}{4}(x-h)^2+k$

Given two points, how can I find an $h$ & $k$ to fit the equation to pass through both of these points?

For example: for points $(0, 0)$ & $(10, 0)$, $y=-\frac{1}{4}(x-5)^2+\frac{25}{4}$, will pass through the two points. However, I got lost on what to do when there's a difference in the $y$-coordinates between the two points like for $(0, 0)$ & $(10,-4)$.

monkeymode
  • 3

1 Answers1

1

You have the points that you can plug into $$ y = a (x-h)^2 + k $$ to get the necessary equations.

For your case $a=-\tfrac{1}{4}$ and

$$\begin{aligned} y_1 & = -\tfrac{1}{4} ( x_1 - h )^2 + k \\ y_2 & = -\tfrac{1}{4} ( x_2 - h )^2 + k \\ \end{aligned}$$

for some pair $(x_1,y_1)$ and $(x_2,y_2)$ of points.

Now if you subtract one from another and expand out the binomial $(x_i -h)^2 = x_i^2 - 2 x_i h + h^2$ you can get the $h^2$ term to cancel which makes the equation solvable for $h$ quite easy

$$ y_2 - y_1 = a \left( 2 h ( x_1 - x_2 ) - x_1^2 - x_2^2 \right)$$

$$ \frac{y_2-y_1}{a} -x_2^2 + x_1^2 = 2 h ( x_1-x_2 )$$

$$ \frac{a ( x_1^2-x_2^2) -y_1 +y_2}{2 a (x_1-x_2)} = h $$

and once $h$ is known, take each one of the original equations to solve for $k$. For example

$$ k = y_1 - a (x_1-h)^2 $$

Fun fact: you can use a similar technique to find the circle that goes through three points.

John Alexiou
  • 13,816
  • This is correct. That said, I wish you had not answered a question where the OP shows no work of their own. They have not even taken advantage of the hint that is essentially your solution. – Ethan Bolker Dec 27 '23 at 01:53
  • 1
    @EthanBolker - I thought the point of this site was to promote learning. I did not provide a numeric answer, and the OP did some work for the first set of points and stumbled for the second set. So I generalized the process in order to promote learning how to think about general problems instead of specific ones. – John Alexiou Dec 27 '23 at 02:00
  • Fair point. I'd have made the other call in this case. – Ethan Bolker Dec 27 '23 at 12:44