Hello, henryhighstudent!

$\displaystyle \text{In hilly areas, radio reception may be poor. \: Consider a situation}$

$\displaystyle \text{where an FM transmitter is located at the point }(-1,1)$

$\displaystyle \text{behind a hill modeled by the graph of: }\,y \:=\:x-x^2$

$\displaystyle \text{and a radio is on the opposite side of the hill.}$

$\displaystyle \text{Assume that the }x\text{-axis represents ground level at the base of the hill.}$

$\displaystyle \text{(a) What is the closest the radio can be to the hill}$

. . . . $\displaystyle \text}so that reception is unobstructed?}$

Code:

T |
o |
(-1,1) o |
| o
| *
| * * o
| * * o
|* * o
| o
- - - - - * - - - - - - * - - - - - - - - o - -
O 1 R

The transmitter is at: .$\displaystyle T(-1,\,1)$

The radio is at $\displaystyle \,R.$

When $\displaystyle \,R$ is at the closest, line $\displaystyle \,T\!R$ is tangent to the parabola

. . but *not* at its vertex.

Let line $\displaystyle T\!R$ have slope $\displaystyle \,m.$

The equation of $\displaystyle T\!R$ is: .$\displaystyle y - 1 \:=\:m(x+1) \quad\Rightarrow\quad y \:=\:mx + (m+1)$ .[1]

Intersection of $\displaystyle T\!R$ and the parabola: .$\displaystyle mx + m + 1 \:=\:x - x^2$

$\displaystyle x^2 + (m\!-\!1)x + (m\!+\!1) \:=\:0 \quad\Rightarrow\quad x \;=\;\dfrac{-(m\!-\!1) \pm\sqrt{(m\!-\!1)^2 - 4(m\!+\!1)}}{2} $

If $\displaystyle T\!R$ is *tangent* to the parabola, there is *one *intersection point;

. . the discriminant is zero.

Hence, we have: .$\displaystyle (m-1)^2 - 4(m_1) \:=\:0 \quad\Rightarrow\quad m^2 - 6m - 3 \:=\:0 $

. . and we have: .$\displaystyle m \:=\:\dfrac{6\pm\sqrt{36+12}}{2} \;=\;3 \pm 2\sqrt{3}$

We see that slope must be negative: .$\displaystyle m \:=\:3 - 2\sqrt{3}$

Substitute into [1].

The equation of $\displaystyle T\!R$ is: .$\displaystyle y \;=\;(3-2\sqrt{3})x + 4-2\sqrt{3}$

We want the $\displaystyle \,x$-intercept of $\displaystyle T\!R.$

. . $\displaystyle (3-2\sqrt{3})x + 4 - 2\sqrt{3} \:=\:0 \quad\Rightarrow\quad x \;=\;\dfrac{-4 + 2\sqrt{3}}{3-2\sqrt{3}} $

which rationalizes to: .$\displaystyle x \;=\;\dfrac{2\sqrt{3}}{3}$

*HA!* . . . Just noticed the wording of the question.

What is the closest the radio can be **to the hill** . . . ?

$\displaystyle \,x$ is the distance from $\displaystyle \,R$ to $\displaystyle \,O.$

The distance from $\displaystyle \,R$ to the hill is: .$\displaystyle \dfrac{2\sqrt{3}}{3} - 1$ units.