# Finding the equation for a parabola given two points and the vertex on a line.

Find the equations for the parabolas $y = a(x-h)^2+k$ that go through the points $(x_0,y_0)$ and $(x_1,y_1)$. The vertex is on the line $y = mx + b$.

I am working with hard values for the points and the line the vertex falls on. I am looking for a strategy to solve this kind of problem. Evidently there are two solutions, so I suspect the solution involves solving a quadratic equation.

The following page describes how to find parabolas through two points. https://www.illustrativemathematics.org/content-standards/tasks/379

Sample problem.

Find the equations for the parabolas that go through the points $(2,5)$ and $(6,53)$. The vertex is on the line $y = 2x + 1$.

I have posted my attempt as an answer.

• To get started, try substituting in both $(x_0,y_0)$ and $(x_1,y_1)$ and solving the simultaneous equations to get a general parabola that passes through both points. – RichoKicked800goals Apr 10 '18 at 1:35
• hint: coordinates of vertex will be $(h, mh+b)$ – Vasya Apr 10 '18 at 1:38
• Re: Kelly, If I am not mistaken, subtracting the equations gives something like $a(x_1^2-x_0^2-2h(x_1-x_0)))-((y_1-y_0)=0$. This appears to be a hyperbola with the variables being $a$ and $h$. I am not sure how to use it in conjunction with $y=mx+b$ to pin down the third points to define the equations for the parabola. – sgeos Apr 10 '18 at 22:59
• Re: Vasya, I think my conceptual gap is using $y=mx+b$ to pin down the third points that define the equations for the parabolas. I need to solve for $a$, $h$ and $k$ given $(x_0,y_0)$, $(x_1,y_1)$, $m$ and $b$. Getting from there to $(h, k=mh+b)$ is where I seem to have a conceptual gap. – sgeos Apr 10 '18 at 23:06
• Solving for $a$ and subtracting to eliminate it gives $\frac{y_0-k}{(x_0-h)^2} = \frac{y_1-k}{(x_1-h)^2}$. Evidently $$h=\frac{2 (x_0 (k-y_1) + x_1 (y_0 - k)) \pm \sqrt{(-2 (x_0 (k-y_1) + x_1 (k-y_0))^2 - 4(y_0-y_1)(x_0^2(k-y_1)+x_1^2(y_0-k))}}{2(y_0-y_1)}$$ and $$k=\frac{h^2(y_1-y_0)+2h(x_1y_0-x_0y_1)+x_0^2y_1-x_1^2y_0}{(x_0-x_1)(-2h+x_0+x_1)}$$ Assuming $x = h$ and $y = k$ on the vertex, $$mh+b=\frac{h^2(y_1-y_0)+2h(x_1y_0-x_0y_1)+x_0^2y_1-x_1^2y_0}{(x_0-x_1)(-2h+x_0+x_1)}$$ Is this the most straightforward way to solve this kind of problem? I am not convinced that it is. – sgeos Apr 10 '18 at 23:37

One way to attack a problem like this is to try to transform it into something that is simpler to solve. For this problem, we might want to try using different coordinate axes.

Rotating the horizontal axis would change how we identify the vertex of the parabola because the minimum or maximum "vertical" value would occur at a different point. Rotating the vertical axis would cause the parabola no longer to be the graph of a function. So let's keep the axes in the same orientation as the given $x$ and $y$ axes. But we can put the origin anywhere we want and scale either axis by any amount independently (possibly even reversing the positive direction) without changing how we identify the vertex.

The problem is relatively easy when the $y$-coordinates of both points are the same, because then $h = \frac{x_1 + x_2}2.$ So let's consider only the case in which $y_1 \neq y_2,$ and suppose we choose coordinate axes labeled $X$ and $Y$ such that the two points the parabola must pass through have $(X,Y)$ coordinates $(-1,-1)$ and $(1,1).$ We can do this by setting \begin{align} X &= \frac{2}{x_2 - x_1}\left(x - \frac{x_1 + x_2}{2}\right), \tag1\\ Y &= \frac{2}{y_2 - y_1}\left(y - \frac{y_1 + y_2}{2}\right). \tag2 \end{align}

Then \begin{align} x &= \left(\frac{x_2 - x_1}{2}\right)X + \frac{x_1 + x_2}{2}, \\ y &= \left(\frac{y_2 - y_1}{2}\right)Y + \frac{y_1 + y_2}{2}, \end{align} and the equation of the line $y = mx + b$ can be rewritten as $$\left(\frac{y_2 - y_1}{2}\right)Y + \frac{y_1 + y_2}{2} = m\left(\frac{x_2 - x_1}{2}\right)X + m\left(\frac{x_1 + x_2}{2}\right) + b,$$ that is, $Y = MX + B,$ where \begin{align} M &= \frac{mx_2 - mx_1}{y_2 - y_1}, \tag3\\ B &= \frac{mx_1 + mx_2 - y_1 - y_2 + 2b}{y_2 - y_1}. \tag4 \end{align}

Let the equation of the parabola in these coordinates be $$Y = A(X - H)^2 + K. \tag5$$ Since the parabola passes through the points $(X,Y) = (-1,-1)$ and $(X,Y) = (1,1),$ we know that \begin{align} -1 &= A(-1 - H)^2 + K = A(1 - 2H + H^2) + K, \\ 1 &= A(1 - H)^2 + K = A(1 + 2H + H^2) + K, \\ -1 - 1 &= (A(1 + 2H + H^2) + K) - (A(1 - 2H + H^2) + K),\\ -2 &= 4AH, \end{align} from which we find that $$A = -\frac{1}{2H}. \tag6$$ Solving for $K$ in Equation $(5),$ $$K = 1 - A(1 - H)^2 = 1 + \frac{(1 - H)^2}{2H} = \frac12\left(H + \frac1H\right). \tag7$$

Therefore the vertex of the parabola has coordinates $\left(H, \frac12\left(H + \frac1H\right)\right).$ In order for that point to lie on the line $Y = MX + B,$ we require that

\begin{align} \frac12\left(H + \frac1H\right) &= MH + B, \\ 0 &= \left(M - \frac12\right) H + B - \frac1{2H} , \\ 0 &= \left(M - \frac12\right) H^2 + BH - \frac12 . \end{align}

In the case where $M = \frac12,$ this equation implies that $H = -\frac{1}{2B}.$ This will lead to exactly one solution if $B\neq 0$; the problem has no solution if $M = \frac12$ and $B = 0.$ But if $M \neq \frac12$ we have a quadratic equation in $H,$ whose solution (if it exists) is $$H = \frac{-B \pm \sqrt{B^2 + 2M - 1}}{2M - 1}. \tag8$$ Note that if $M < \frac12$ there is a solution only if $\lvert B\rvert \geq \sqrt{1 - 2M},$ and there is only one solution when $M < \frac12$ and $\lvert B\rvert = \sqrt{1 - 2M}$; otherwise there are two solutions.

Now to apply this to the general problem, given the coordinates $(x_1,y_1)$ and $(x_2,y_2)$ and the equation of the line $y = mx + b,$ we can take Equation $(5)$ and use Equations $(1)$ and $(2)$ to substitute for $X$ and $Y,$ obtaining $$\frac{2}{y_2 - y_1}\left(y - \frac{y_1 + y_2}{2}\right) = A\left(\frac{2}{x_2 - x_1} \left(x - \frac{x_1 + x_2}{2}\right) - H\right)^2 + K.$$

Redistributing some factors, we get $$\frac{2}{y_2 - y_1}\left(y - \frac{y_1 + y_2}{2}\right) = \frac{4A}{(x_2 - x_1)^2} \left(x - \frac{x_1 + x_2}{2} - \frac{x_2 - x_1}{2}H\right)^2 + K$$ and then $$y - \frac{y_1 + y_2}{2} = \frac{2A(y_2 - y_1)}{(x_2 - x_1)^2} \left(x - \frac{x_1 + x_2}{2} - \frac{x_2 - x_1}{2}H\right)^2 + \frac{K(y_2 - y_1)}{2}$$ and finally, by adding a constant to both sides, $$y = \frac{2A(y_2 - y_1)}{(x_2 - x_1)^2} \left(x - \frac{x_1 + x_2}{2} - \frac{x_2 - x_1}{2}H\right)^2 + \frac{K(y_2 - y_1) + y_1 + y_2}{2}. \tag9$$

At this point you should be able to recognize which of these expressions is $a,$ $h,$ and $k$ in the desired equation $y = a(x - h)^2 + k.$

To write this completely in terms of the variables $x$ and $y$ and the parameters $x_1,$ $y_1,$ $x_2,$ $y_2,$ $m,$ and $b$ given in the problem statement, you could use Equations $(6)$ and $(7)$ to substitute for $A$ and $K,$ then use Equation $(8)$ to substitute for $H,$ and finally use Equations $(3)$ and $(4)$ to substitute for $M$ and $B.$ I would not recommend writing this result out in full detail, however; it seems more practical to me to say the equation of the parabola is Equation $(9)$ where $A,$ $H,$ and $K$ are defined by the equations above. This enables plugging in actual numeric values to work examples.

In fact, I would even recommend introducing some additional symbols in order to make Equation $(9)$ easier to work with: \begin{align} \bar x &= \frac{x_1 + x_2}{2}, & \delta_x &= \frac{x_2 - x_1}{2},\\ \bar y &= \frac{y_1 + y_2}{2}, & \delta_y &= \frac{y_2 - y_1}{2}. \end{align} Then Equation $(9)$ becomes $$y = \frac{A\delta_y}{\delta_x^2}\left(x - \bar x - H\delta_x\right)^2 + K\delta_y + \bar y.$$

• Thank you much. I will work through this as time permits. – sgeos Apr 19 '18 at 4:29
• $\frac{2}{x_2-x_1}$ scales the axis so the distance between $x_1$ and $x_2$ is exactly 2. $-\frac{x_1+x_2}{2}$ is an offset that centers the axis between $x_1$ and $x_2$. That makes sense. Geometrically, what does $M=\frac{1}{2}$ imply? What is the implication when $B=0$ and $M=\frac{1}{2}$? What do the additional terms you introduced represent conceptually? – sgeos Apr 20 '18 at 5:02
• The terms $\bar x, \delta_x, \bar y,$ and $\delta_y$ are just the offsets and scaling factors (or their reciprocals) that scale and shift the points $(x_1,y_2)$ and $(x_2,y_2)$ to $(-1,-1)$ and $(1,1).$ The capital letters correspond to the lower-case letters but in the transformed coordinates instead of the original coordinates. The graph of $K=\frac12(H+(1/H))$ has two branches, which are asymptotic to the $Y$ axis (halfway between the two point) and to the line $Y = \frac12X$--that is, the line with slope $M=\frac12$ in transformed coordinates. – David K Apr 20 '18 at 5:34
• Are the asymptotic properties of $K=\frac{1}{2}(H+\frac{1}{H})$ an artifact of the scaling or a property of the original problem? – sgeos Apr 20 '18 at 6:03
• The graph is a hyperbola. When you transform back to the original coordinates, it's still a hyperbola with one vertical asymptote and one sloped asymptote, but the intersection point of the asymptotes will be $(\bar x,\bar y)$ instead of $(0,0)$ and the slope of the sloped asymptote will be $\frac12\delta_y/\delta_x$ instead of $\frac12.$ – David K Apr 20 '18 at 11:47

This is my attempt to answer my own question.

Substituting for $x$ and $y$, subtracting equations and solving for $a$ gives the following. $$y_1 - y_0=(a(x_1-h)^2+k) - (a(x_0-h)^2+k) \\ (y_1 - y_0)=a((x_1-h)^2 - (x_0-h)^2) \\ a = \frac{y_1 - y_0}{(x_1-h)^2 - (x_0-h)^2}$$

Substituting for $x$, $y$ and $a$, adding equations and solving for $k$ gives the following. $$y_1 + y_0=(a(x_1-h)^2+k) + (a(x_0-h)^2+k) \\ (y_1 + y_0)=a((x_1-h)^2 + (x_0-h)^2) + 2k \\ (y_1 + y_0)=\frac{y_1 - y_0}{(x_1-h)^2 - (x_0-h)^2}((x_1-h)^2 + (x_0-h)^2) + 2k \\ 2k = (y_1 + y_0) - \frac{(y_1 - y_0)((x_1-h)^2 + (x_0-h)^2)}{(x_1-h)^2 - (x_0-h)^2} \\ 2k = \frac{(y_1 + y_0)((x_1-h)^2 - (x_0-h)^2) - (y_1 - y_0)((x_1-h)^2 + (x_0-h)^2)}{(x_1-h)^2 - (x_0-h)^2} \\ 2k = \frac{(y_0(x_1-h)^2+y_1(x_1-h)^2-y_0(x_0-h)^2-y_1(x_0-h)^2) - (-y_0(x_1-h)^2+y_1(x_1-h)^2-y_0(x_0-h)^2+y_1(x_0-h)^2)}{(x_1-h)^2 - (x_0-h)^2} \\ 2k = \frac{(y_0(x_1-h)^2+y_1(x_1-h)^2-y_0(x_0-h)^2-y_1(x_0-h)^2) + (y_0(x_1-h)^2-y_1(x_1-h)^2+y_0(x_0-h)^2-y_1(x_0-h)^2)}{(x_1-h)^2 - (x_0-h)^2} \\ 2k = \frac{y_0(x_1-h)^2+y_0(x_1-h)^2+y_1(x_1-h)^2-y_1(x_1-h)^2-y_0(x_0-h)^2+y_0(x_0-h)^2-y_1(x_0-h)^2-y_1(x_0-h)^2}{(x_1-h)^2 - (x_0-h)^2} \\ 2k = \frac{2(y_0(x_1-h)^2 - y_1(x_0-h)^2)}{(x_1-h)^2 - (x_0-h)^2} \\ k = \frac{y_0(x_1-h)^2 - y_1(x_0-h)^2}{(x_1-h)^2 - (x_0-h)^2} \\$$

Substituting for $a$ and $k$ in the original equation gives a parabola in terms of $h$ that goes through the points $(x_0,y_0)$ and $(x_1,y_1)$.

$$y = a(x-h)^2+k \\ y = \frac{y_1 - y_0}{(x_1-h)^2 - (x_0-h)^2}(x-h)^2 + \frac{y_0(x_1-h)^2 - y_1(x_0-h)^2}{(x_1-h)^2 - (x_0-h)^2} \\ y = \frac{(y_1 - y_0)(x-h)^2 + y_0(x_1-h)^2 - y_1(x_0-h)^2}{(x_1-h)^2 - (x_0-h)^2} \\ y = \frac{(y_1 - y_0)(x-h)^2 + y_0(x_1-h)^2 - y_1(x_0-h)^2}{(x_1^2-2x_1h+h^2) - (x_0^2-2x_0h+h^2)} \\ y = \frac{(y_1 - y_0)(x-h)^2 + y_0(x_1-h)^2 - y_1(x_0-h)^2}{x_1^2-x_0^2-2x_1h+2x_0h} \\ y = \frac{(y_1 - y_0)(x-h)^2 + y_0(x_1-h)^2 - y_1(x_0-h)^2}{(x_1-x_0)(x_0+x_1)-2h(x_1-x_0)} \\ y = \frac{(y_1 - y_0)(x-h)^2 + y_0(x_1-h)^2 - y_1(x_0-h)^2}{(x_1-x_0)(x_0+x_1-2h)} \\ y = \frac{(y_1 - y_0)(x-h)^2 + y_0(x_1-h)^2 - y_1(x_0-h)^2}{(x_0-x_1)(2h-x_0-x_1)}$$

The above is clearly useless when $x_0=x_1$ or $y_0=y_1$. In any case, the vertex is located at $(h,mh+b)$. Substituting for $x$, $y$ and solving for $h$ gives the following.

$$y = a(x-h)^2+k \\ mh+b = a(h-h)^2+k \\ mh+b = k \\ h = \frac{k}{m} - \frac{b}{m} \\ h = \frac{y_0(x_1-h)^2 - y_1(x_0-h)^2}{m((x_1-h)^2 - (x_0-h)^2)} - \frac{b}{m} \\ h = \frac{y_0(x_1-h)^2 - y_1(x_0-h)^2 - b((x_1-h)^2 - (x_0-h)^2)}{m((x_1-h)^2 - (x_0-h)^2)} \\ hm((x_1-h)^2 - (x_0-h)^2) = y_0(x_1-h)^2 - y_1(x_0-h)^2 - b((x_1-h)^2 - (x_0-h)^2) \\ hm((x_1-h)^2 - (x_0-h)^2) + b((x_1-h)^2 - (x_0-h)^2) = y_0(x_1-h)^2 - y_1(x_0-h)^2 \\ (hm + b)((x_1-h)^2 - (x_0-h)^2) = y_0(x_1-h)^2 - y_1(x_0-h)^2 \\ 0 = (hm + b)((x_1-h)^2 - (x_0-h)^2) + y_1(x_0-h)^2 - y_0(x_1-h)^2 \\ 0 = 2 h^2 m x_0 - 2 h^2 m x_1 + h^2 y_1 - h^2 y_0 + h m x_1^2 - h m x_0^2 + 2 h x_1 y_0 - 2 h x_0 y_1 + 2 b h x_0 - 2 b h x_1 - b x_0^2 + b x_1^2 + x_0^2 y_1 - x_1^2 y_0 \\ 0 = h^2 ( 2 m (x_0 - x_1) + y_1 - y_0 ) + h ( m (x_1^2 - x_0^2) + 2 (x_1 (y_0 - b) - x_0 (y_1 - b)) ) + b (x_1^2 - x_0^2) + x_0^2 y_1 - x_1^2 y_0$$

The quadratic equation can now be used to solve for $h$.

$$h = \frac{-B\pm\sqrt{B^2-4 A C}}{2A} \\ A = 2 m (x_0 - x_1) + y_1 - y_0 \\ B = m (x_1^2 - x_0^2) + 2 (x_1 (y_0 - b) - x_0 (y_1 - b)) \\ C = b (x_1^2 - x_0^2) + x_0^2 y_1 - x_1^2 y_0 \\ h = \frac{-( m (x_1^2 - x_0^2) + 2 (x_1 (y_0 - b) - x_0 (y_1 - b))) \pm\sqrt{( m (x_1^2 - x_0^2) + 2 (x_1 (y_0 - b) - x_0 (y_1 - b)))^2-4 ( 2 m (x_0 - x_1) + y_1 - y_0 ) (b (x_1^2 - x_0^2) + x_0^2 y_1 - x_1^2 y_0)}}{2( 2 m (x_0 - x_1) + y_1 - y_0 )} \\ h = \frac{- m (x_1^2 - x_0^2) - 2 (x_1 (y_0 - b) - x_0 (y_1 - b)) \pm\sqrt{ (x_0 - x_1)^2 (2 m x_0 (2 b + m x_1 - 2 y_1) + 4 m x_1 (b - y_0) + 4 (b - y_0) (b - y_1) + m^2 x_0^2 + m^2 x_1^2)}}{2( 2 m (x_0 - x_1) + y_1 - y_0 )} \\ h = \frac{- m (x_1^2 - x_0^2) - 2 (x_1 (y_0 - b) - x_0 (y_1 - b)) \pm (x_0 - x_1) \sqrt{ 2 m x_0 (2 b + m x_1 - 2 y_1) + 4 m x_1 (b - y_0) + 4 (b - y_0) (b - y_1) + m^2 x_0^2 + m^2 x_1^2}}{2( 2 m (x_0 - x_1) + y_1 - y_0 )} \\ h = \frac{- m (x_1^2 - x_0^2) - 2 (x_1 (y_0 - b) - x_0 (y_1 - b)) \pm (x_0 - x_1) \sqrt{ (2 b + m ( x_0 + x_1 ) - 2 y_1)^2 - 4 (y_0 - y_1) (b + m x_1 - y_1) }}{2( 2 m (x_0 - x_1) + y_1 - y_0 )} \\$$

Solution to sample problem.

$$x_0 = 2 \\ y_0 = 5 \\ x_1 = 6 \\ y_1 = 53 \\ m = 2 \\ b = 1 \\$$ $$h = \frac{- m (x_1^2 - x_0^2) - 2 (x_1 (y_0 - b) - x_0 (y_1 - b)) \pm (x_0 - x_1) \sqrt{ (2 b + m ( x_0 + x_1 ) - 2 y_1)^2 - 4 (y_0 - y_1) (b + m x_1 - y_1) }}{2( 2 m (x_0 - x_1) + y_1 - y_0 )} \\ h = \frac{- 2 (6^2 - 2^2) - 2 (6 (5 - 1) - 2 (53 - 1)) \pm (2 - 6) \sqrt{ (2 (1) + 2 ( 2 + 6 ) - 2 (53))^2 - 4 (5 - 53) (1 + 2 (6) - 53) }}{2( 2 (2) (2 - 6) + 53 - 5 )} \\ h = \frac{96 \mp 4 \sqrt{ 64 }}{64} = \frac{3 \mp 1}{2} \\ h_0 = 1 \\ h_1 = 2 \\$$

$$k = \frac{y_0(x_1-h)^2 - y_1(x_0-h)^2}{(x_1-h)^2 - (x_0-h)^2} \\ k = \frac{5(6-h)^2 - 53(2-h)^2}{(6-h)^2 - (2-h)^2} \\ k_0 = \frac{5(6-h_0)^2 - 53(2-h_0)^2}{(6-h_0)^2 - (2-h_0)^2} \\ k_0 = \frac{5(6-1)^2 - 53(2-1)^2}{(6-1)^2 - (2-1)^2} = \frac{72}{24} \\ k_0 = 3\\ k_1 = \frac{5(6-h_1)^2 - 53(2-h_1)^2}{(6-h_1)^2 - (2-h_1)^2} \\ k_1 = \frac{5(6-2)^2 - 53(2-2)^2}{(6-2)^2 - (2-2)^2} = \frac{80}{16} \\ k_1 = 5$$

$$a = \frac{y_1 - y_0}{(x_1-h)^2 - (x_0-h)^2} \\ a = \frac{53 - 5}{(6-h)^2 - (2-h)^2} \\ a_0 = \frac{48}{(6-h_0)^2 - (2-h_0)^2} \\ a_0 = \frac{48}{(6-1)^2 - (2-1)^2} = \frac{48}{24} \\ a_0 = 2 \\ a_1 = \frac{48}{(6-h_1)^2 - (2-h_1)^2} \\ a_1 = \frac{48}{(6-2)^2 - (2-2)^2} = \frac{48}{16} \\ a_1 = 3$$

$$y = a(x-h)^2+k \\ y = 2(x-1)^2+3 \\ y = 3(x-2)^2+5$$