Key Concepts and Formulas

• Parametric Equation of a Parabola: For the parabola $y^2 = 4ax$, a point on the parabola can be represented as $(at^2, 2at)$.
• Equation of Normal to a Parabola: The equation of the normal to the parabola $y^2 = 4ax$ at the point $(at^2, 2at)$ is given by $y = -tx + 2at + at^3$. If the slope of normal is $m$, then $y = mx + 2a(-m) + a(-m)^3$. So $y = mx - 2am - am^3$.
• Distance Formula: The distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.
• Shortest Distance: The shortest distance between a curve and a circle is along the normal to the curve that passes through the center of the circle, minus the radius of the circle.

Step-by-Step Solution

Step 1: Analyze the Parabola

The equation of the parabola is $y^2 = 8x$. We want to express it in the standard form $y^2 = 4ax$. $y^2 = 8x = 4(2)x$ Thus, $a = 2$. A point on the parabola can be represented as $P(at^2, 2at) = P(2t^2, 4t)$. The parameter $t$ is related to the slope of the tangent at the point.

Step 2: Find the Equation of the Normal to the Parabola

The equation of the normal at any point $(at^2,2at)$ to the parabola $y^2=4ax$ is $y = -tx + 2at + at^3$. Since $a=2$, the normal equation becomes $y = -tx + 2(2)t + 2t^3$ $y = -tx + 4t + 2t^3$

Now, let the slope of the normal be $m = -t$ so $t = -m$. Substituting $t=-m$ into the parametric point $P(2t^2, 4t)$ gives $P(2m^2, -4m)$. Substituting $t = -m$ into the normal equation, we get: $y = -(-m)x + 4(-m) + 2(-m)^3$ $y = mx - 4m - 2m^3$

Step 3: Analyze the Circle

The equation of the circle is $x^2 + y^2 + 12y + 35 = 0$. To find its center and radius, we complete the square: $x^2 + (y^2 + 12y) + 35 = 0$ $x^2 + (y^2 + 12y + 36) - 36 + 35 = 0$ $x^2 + (y + 6)^2 = 1$ This is in the standard form $(x - h)^2 + (y - k)^2 = r^2$. The center of the circle is $C(0, -6)$ and the radius is $r = 1$.

Step 4: Find the Normal Passing Through the Center of the Circle

The shortest distance is along the normal to the parabola that passes through the center of the circle. Substitute the coordinates of the center $C(0, -6)$ into the equation of the normal $y = mx - 4m - 2m^3$: $-6 = m(0) - 4m - 2m^3$ $-6 = -4m - 2m^3$ $2m^3 + 4m - 6 = 0$ $m^3 + 2m - 3 = 0$ By inspection, we see that $m = 1$ is a root: $(1)^3 + 2(1) - 3 = 1 + 2 - 3 = 0$. So, $m = 1$. We can perform polynomial division $(m^3 + 2m - 3) \div (m-1)$ which results in $m^2 + m + 3$. The discriminant of this quadratic is $1^2 - 4(1)(3) = 1 - 12 = -11 < 0$. Therefore, there are no other real roots.

Step 5: Find the Point on the Parabola Closest to the Circle

The point on the parabola is $P(2m^2, -4m)$. Substituting $m = 1$: $P(2(1)^2, -4(1)) = P(2, -4)$

Step 6: Calculate the Shortest Distance

The shortest distance is the distance from the point $P(2, -4)$ to the center of the circle $C(0, -6)$, minus the radius $r = 1$. The distance between $P(2, -4)$ and $C(0, -6)$ is: $PC = \sqrt{(2 - 0)^2 + (-4 - (-6))^2} = \sqrt{2^2 + 2^2} = \sqrt{8} = 2\sqrt{2}$ The shortest distance is $PC - r = 2\sqrt{2} - 1$.

Common Mistakes & Tips

• Remember to subtract the radius of the circle from the distance between the point on the curve and the center of the circle.
• The parametric form of the normal simplifies calculations significantly.
• Always double-check your algebra, especially when dealing with cubic equations.

Summary

We determined the equation of the normal to the parabola in terms of its slope, found the normal passing through the center of the circle, and then calculated the distance between the corresponding point on the parabola and the center of the circle. Subtracting the radius of the circle gave the shortest distance between the curves.

The final answer is $\boxed{2\sqrt{2}-1}$, which corresponds to option (B).