In the diagram, $A B C D$ is a square. Points $A(1,4)$ and $B$ are on a parabola that is tangent to the $x$-axis. If $C$ has coordinates $\left(\frac{39}{4}, \frac{37}{4}\right)$, determine the equation of the parabola.

![](https://cdn.mathpix.com/cropped/2024_04_30_22929074a9efc78b4168g-2.jpg?height=658&width=636&top_left_y=877&top_left_x=796)
Since the parabola opens upwards and has only one $x$-intercept, then its equation is of the form $y=a(x-r)^{2}$ for some real numbers $a, r>0$.

Since the point $A(1,4)$ is on the parabola, then $(x, y)=(1,4)$ satisfies the equation of the parabola, which gives $4=a(1-r)^{2}$.

We will obtain a second equation for $a$ and $r$ by determining the coordinates of $B$.

Since $A B C D$ is a square, its diagonals $A C$ and $B D$ cross at their common midpoint, $M$, and are perpendicular at $M$.

Since $A$ has coordinates $(1,4)$ and $C$ has coordinates $\left(\frac{39}{4}, \frac{37}{4}\right)$, then the coordinates of $M$ are $\left(\frac{1}{2}\left(1+\frac{39}{4}\right), \frac{1}{2}\left(4+\frac{37}{4}\right)\right)=\left(\frac{43}{8}, \frac{53}{8}\right)$.

To get from $A$ to $M$, we go $\frac{35}{8}$ units right and $\frac{21}{8}$ units up.

Thus, to get from $M$ to $B$, we go $\frac{21}{8}$ units right and $\frac{35}{8}$ units down.

This means that the coordinates of $B$ are $\left(\frac{43}{8}+\frac{21}{8}, \frac{53}{8}-\frac{35}{8}\right)=\left(8, \frac{9}{4}\right)$.

To see why this "over and up" approach is correct, draw a vertical line from $M$ to $P$ on the $x$-axis, and draw horizontal lines from $A$ and $B$ to points $S$ and $T$, respectively, on $M P$.

![](https://cdn.mathpix.com/cropped/2024_04_30_75c1cf70fbe3735bfd92g-04.jpg?height=651&width=635&top_left_y=954&top_left_x=802)

Since $A S$ and $B T$ are perpendicular to $M P$, then $\triangle A S M$ and $\triangle M T B$ are right-angled.

Note that $A M=M B$, since $A B C D$ is a square and $M$ is its centre.

Since $A M$ and $M B$ are perpendicular, then $\angle A M S+\angle T M B=90^{\circ}$.

But $\angle S A M+\angle A M S=90^{\circ}$, so $\angle S A M=\angle T M B$.

This means that $\triangle A S M$ and $\triangle M T B$ have equal corresponding angles.

Since $A M=M B$, these triangles are congruent, and so $A S=M T=\frac{35}{8}$ and $T B=S M=\frac{21}{8}$. Thus, the coordinates of $B$ are indeed $\left(8, \frac{9}{4}\right)$.

Since $B$ is also on the parabola, then $\frac{9}{4}=a(8-r)^{2}$.

Dividing this equation by the first equation $4=a(1-r)^{2}$, we obtain $\frac{9}{16}=\left(\frac{8-r}{1-r}\right)^{2}$.

Since $1<r<8$ (the vertex of the parabola is between $A$ and $B$ ), then $\frac{8-r}{1-r}<0$ and so $\frac{8-r}{1-r}=-\frac{3}{4}$.

Thus, $32-4 r=3 r-3$ and so $7 r=35$ or $r=5$.

Since $4=a(1-r)^{2}$, then $4=a(1-5)^{2}$ and so $a=\frac{4}{16}=\frac{1}{4}$.

Therefore, the equation of the parabola is $y=\frac{1}{4}(x-5)^{2}$.