Question

find shortest distance between (x-15)^2 + y^2 = 4 and y = x^2/20 by equating their slopes

Answer 1

5\sqrt{2}-2

Answer 2

To find the shortest distance between the circle $(x-15)^2 + y^2 = 4$ and the parabola $y = x^2/20$, we use the principle that the shortest distance between two curves lies along their common normal. For a circle, any normal passes through its center.

1. Identify the properties of the curves:

   • The circle has equation $(x-15)^2 + y^2 = 4$. Its center is $C(15, 0)$ and its radius is $r = 2$.
   • The parabola has equation $y = x^2/20$.

2. Condition for shortest distance:

   Let $P_1(x_1, y_1)$ be the point on the parabola and $P_2(x_2, y_2)$ be the point on the circle such that the distance $P_1P_2$ is minimized. The line segment $P_1P_2$ must be perpendicular to the tangents of both curves at these points (i.e., it must be a common normal). Since $P_1P_2$ is a normal to the circle, it must pass through the center of the circle $C(15,0)$. Therefore, $P_1$, $P_2$, and $C$ are collinear. This implies that the normal to the parabola at $P_1$ must pass through the center of the circle $C(15,0)$.

3. Find the slope of the tangent and normal to the parabola:

   For the parabola $y = x^2/20$, differentiate with respect to $x$: $\frac{dy}{dx} = \frac{2x}{20} = \frac{x}{10}$.

   The slope of the tangent at $P_1(x_1, y_1)$ is $m_t = \frac{x_1}{10}$. The slope of the normal at $P_1(x_1, y_1)$ is $m_N = -\frac{1}{m_t} = -\frac{10}{x_1}$.

4. Equate the slope of the normal to the slope of the line $CP_1$:

   The line $CP_1$ passes through $C(15,0)$ and $P_1(x_1, y_1)$. Its slope is $\frac{y_1 - 0}{x_1 - 15} = \frac{y_1}{x_1 - 15}$. Since the normal to the parabola at $P_1$ passes through $C$, we equate their slopes: $\frac{y_1}{x_1 - 15} = -\frac{10}{x_1}$

   $x_1 y_1 = -10(x_1 - 15)$

   $x_1 y_1 = -10x_1 + 150 \quad (*)$

5. Solve for $P_1(x_1, y_1)$:

   Since $P_1(x_1, y_1)$ lies on the parabola, $y_1 = x_1^2/20$. Substitute this into equation $(*)$: $x_1 \left(\frac{x_1^2}{20}\right) = -10x_1 + 150$

   $\frac{x_1^3}{20} = -10x_1 + 150$

   $x_1^3 = -200x_1 + 3000$

   $x_1^3 + 200x_1 - 3000 = 0$

   By inspection, we can test integer factors of 3000. Let's try $x_1 = 10$: $10^3 + 200(10) - 3000 = 1000 + 2000 - 3000 = 0$. So, $x_1 = 10$ is a root. Dividing the cubic equation by $(x_1-10)$ gives $x_1^2 + 10x_1 + 300 = 0$. The discriminant of this quadratic is $\Delta = 10^2 - 4(1)(300) = 100 - 1200 = -1100 < 0$. Thus, there are no other real roots. So, $x_1 = 10$ is the unique real solution. Now find $y_1$: $y_1 = \frac{x_1^2}{20} = \frac{10^2}{20} = \frac{100}{20} = 5$. The point on the parabola closest to the circle is $P_1(10, 5)$.

6. Calculate the distance from the center of the circle to $P_1$:

   Distance $CP_1 = \sqrt{(10-15)^2 + (5-0)^2} = \sqrt{(-5)^2 + 5^2} = \sqrt{25 + 25} = \sqrt{50} = 5\sqrt{2}$.

7. Calculate the shortest distance:

   The point $P_1(10,5)$ is outside the circle, because its distance from the center $C(15,0)$ is $5\sqrt{2} \approx 7.07$, which is greater than the radius $r=2$. The shortest distance between the parabola and the circle is the distance $CP_1$ minus the radius of the circle. Shortest distance $= CP_1 - r = 5\sqrt{2} - 2$.