What is another, perhaps quicker and nicer way of solving this question? In a calculus book, a question reads:

A car is traveling at night along a highway shaped like a parabola with its vertex at the origin. The car starts at a point 100 m west and 100 m north of the origin and travels in an easterly direction. There is a statue located 100 m east and 50 m north of the origin. At what point on the highway will the car's headlights illuminate the statue?

I want to know how you would solve this problem, because the method I am using is very round-about. There must be a much more logical way of solving this problem, and I hope you will share it with me. Thank you.

This is my solution..
The function appears to be of the form $$y = f(x) = ax^2$$
$$100 = a(100)^2$$
$$a = 1/100$$
$$y = \frac{x^2}{100}$$
Then I decided (by solving a few tangent equations and generalizing) that the tangent line expression at a point $(p, f(x))$ is given by the function:
$$T(p) = T_m(p)*x + T_b(p)$$
Where $T_m(p)$ represents the slope for a point, $p$ on $f(x)$ and $T_b(p)$ represents the y-intercept value.
$$T_m(p) = \frac{df(p)}{dx} =\frac{p}{50}$$
$$T_b(p) = f(p) - \frac{df(p)}{dx} * p = -\frac{p^2}{100}$$
So given that the statue rests on $(100, 50)$ we know the equation of this tangent line outputs $50$ when $100$ is input.
$$T(p) = \frac{p}{50}*x - \frac{p^2}{100}$$
$$50 = \frac{p}{50}*100 - \frac{p^2}{100}$$
the $x$ position of the car is at a point $p$ such that $p$ satisfies the equation.
I end up getting the positive result $x = 100 - 50\sqrt{2}$
 A: Your solution is great. Here's my approach.
We know that the parabola has equation $f(x)=\dfrac{x^2}{100}$. Now suppose that the car illuminates the statue when $x=p$. Then the car's location must be at $\left(p,\frac{p^2}{100}\right)$, and the line connecting this point to $(100,50)$ must have a slope of $f'(p)=\dfrac{p}{50}$. Hence, using the slope formula, we obtain:
$$ \begin{align*}
\dfrac{\frac{p^2}{100}-50}{p-100} &= \dfrac{p}{50} \\
\dfrac{p^2-5000}{100(p-100)} &= \dfrac{p}{50} \\
\dfrac{p^2-5000}{2(p-100)} &= p \\
p^2-5000 &= 2p(p-100) \\
p^2-5000 &= 2p^2-200p \\
0 &= p^2-200p+5000 \\
p &= 100 \pm 50\sqrt2 \\
\end{align*} $$
Hence, since we know that the car travels from left to right, we reject the larger solution and conclude that the car illuminates the statue at $x=100-50\sqrt2$.
A: A coworker came to me with this problem today, and I found the solution a slightly different way. My approach wasn’t any simpler, but I still thought I would share:
As both of your approaches demonstrated, $f(x)=\frac{x^2}{100}$ and $f'(x)=\frac{x}{50}$. We also know that the tangent line intersects $(100,50)$. So the tangent line can be defined as:
$$y=mx+b$$
$$50=\frac{x}{50}100+b$$
$$b=50-2x$$
$$y=\frac{x}{50}x+50-2x$$
$$y=\frac{x^2}{50}-2x+50$$
Solve the system: $y=\frac{x^2}{100}$ and $y=\frac{x^2}{50}-2x+50$.
$$\frac{x^2}{100}=\frac{x^2}{50}-2x+50$$
$$\frac{x^2}{100}-2x+50=0$$
$$x^2-200x+5000=0$$
$$x=100±50\sqrt{2}$$
I ignored that value where $x>100$, so the car's location at which its headlight's illuminate the statue is:$$(100-50\sqrt{2}, 150-100\sqrt{2})$$
