# 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 $x = 100 - 50\sqrt{2}$

• Note that the problem expects you to assume that the parabola has axis facing straight up. There are piles of parabolas with vertex the origin that face in all sorts of directions. Aug 9 '13 at 3:18

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 $x=100-50\sqrt2$.

• I really like the way you approach it, always nice to see things from a more focused perspective. Aug 9 '13 at 5:24
• @Adriano sorry if this is late but why can you say "Suppose the car illuminates the statue at point P" could you explain that part? Feb 13 '16 at 5:55
• @dydxx: Certainly we know that such a point $p$ exists, since the question tells us so. Some people like leaving the variable as $x$, but I prefer to use a different variable because I generally think of $x$ as a variable that could change over time, whereas I generally think of other variables such as $p$ as constants that refer to specific, fixed points in time. Feb 13 '16 at 10:58

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:$$(100-50\sqrt{2}, 150-100\sqrt{2})$$