A line with slope 1 passes through point $P$ on the negative $x$-axis and intersects the parabola $y=x^{2}$ at points $Q$ and $R$, as shown. If $P Q=Q R$, then the $y$-intercept of $P R$ is closest to
(A) 9.9
(B) 10.2
(C) 8.2
(D) 9.3
(E) 8.6

![](https://cdn.mathpix.com/cropped/2024_04_20_ac36362783317e0251fdg-101.jpg?height=349&width=423&top_left_y=777&top_left_x=1290)
Suppose that the $y$-intercept of $P R$ is $b$ with $b>0$.

Since $P R$ has slope 1 , then $P R$ has $x$-intercept $-b$, so the coordinates of $P$ are $(-b, 0)$.

Since $P, Q$ and $R$ lie on a line with $P Q=Q R$, then the horizontal distance from $P$ to $Q$ equals the horizontal distance to $Q$ to $R$. In other words, the difference between the $x$-coordinates of $Q$ and $P$ equals that of $R$ and $Q$.

Since $P R$ has slope 1 and $y$-intercept $b$, it has equation $y=x+b$.

We can determine the $x$-coordinates of $Q$ and $R$ by determining the points of intersection of $y=x^{2}$ and $y=x+b$, which we get by solving

$$
\begin{aligned}
x^{2} & =x+b \\
x^{2}-x-b & =0 \\
x & =\frac{1 \pm \sqrt{1-4(1)(-b)}}{2} \quad \text { (by the quadratic formula) } \\
x & =\frac{1 \pm \sqrt{1+4 b}}{2}
\end{aligned}
$$

From the diagram, the $x$-coordinate of $Q$ is $\frac{1-\sqrt{1+4 b}}{2}$ and the $x$-coordinate of $R$ is $\frac{1+\sqrt{1+4 b}}{2}$.

From the given information,

$$
\begin{aligned}
\frac{1-\sqrt{1+4 b}}{2}-(-b) & =\frac{1+\sqrt{1+4 b}}{2}-\frac{1-\sqrt{1+4 b}}{2} \\
\frac{1-\sqrt{1+4 b}}{2}+b & =\sqrt{1+4 b} \\
1-\sqrt{1+4 b}+2 b & =2 \sqrt{1+4 b} \\
1+2 b & =3 \sqrt{1+4 b} \\
(1+2 b)^{2} & =(3 \sqrt{1+4 b})^{2} \\
1+4 b+4 b^{2} & =9(1+4 b) \\
4 b^{2}-32 b-8 & =0 \\
b^{2}-8 b-2 & =0 \\
b & =\frac{8 \pm \sqrt{8^{2}-4(1)(-2)}}{2} \\
b & =\frac{8 \pm \sqrt{72}}{2} \\
b & =4 \pm 3 \sqrt{2}
\end{aligned}
$$

Since $b>0$, then $b=4+3 \sqrt{2} \approx 8.243$.

(In this problem, it is in fact possible to determine which of the given answers is correct using a carefully drawn scale diagram.)

ANSWER: (C)