Answer
The car's headlights will illuminate the statue when the car is at the position 29.29 meters east and 8.58 meters north of the origin.
Work Step by Step
If the parabola's vertex is at the origin, then the equation of the parabola is $y = ax^2$
We can find $a$:
$y = ax^2$
$100 = a(-100)^2$
$a = \frac{1}{100}$
The equation of the parabola is $y = \frac{x^2}{100}$
$y' = \frac{x}{50}$
Let $(x_0, \frac{x_0^2}{100})$ be the point on the parabola where the headlights illuminate the statue. Then the line connecting $(x_0, \frac{x_0^2}{100})$ and the statue's point $(100, 50)$ is a tangent line to the parabola. Thus the slope of this line is equal to the slope of the parabola at $x_0$.
We can equate the slopes to find $x_0$:
$\frac{x_0}{50} = \frac{50-\frac{x_0^2}{100}}{100-x_0}$
$100~x_0-x_0^2 = 2500-\frac{x_0^2}{2}$
$200~x_0-2x_0^2 = 5000-x_0^2$
$x_0^2-200~x_0 + 5000 = 0$
We can use the quadratic formula:
$x_0 = \frac{200\pm \sqrt{(-200)^2-(4)(1)(5000)}}{2(1)}$
$x_0 = \frac{200\pm \sqrt{20,000}}{2}$
$x_0 = 100\pm 50 \sqrt{2}$
There are two solutions so we will choose the solution that is closer to the origin since this is the point where the front of the car is pointing at the statue.
$x_0 = 100 - 50 \sqrt{2}$
$x_0 = 29.29~m$
When $x_0 = 100- 50 \sqrt{2}$:
$y = \frac{(100- 50 \sqrt{2})^2}{100}$
$y = \frac{15,000- 10,000 \sqrt{2}}{100}$
$y = 150- 100 \sqrt{2}$
$y = 8.58~m$
The car's headlights will illuminate the statue when the car is at the position 29.29 meters east and 8.58 meters north of the origin.
