Key Concepts and Formulas

• The shortest distance between a curve and a line occurs where the tangent to the curve is parallel to the line.
• The derivative of a function, $\frac{dy}{dx}$, gives the slope of the tangent line to the curve at a given point.
• The normal to a curve at a point is perpendicular to the tangent at that point. If the slope of the tangent is $m$, the slope of the normal is $-\frac{1}{m}$.

Step-by-Step Solution

Step 1: Determine the slope of the given line.

We are given the line $y = 3x - 3$. Our goal is to find the slope of this line because the tangent to the curve at the closest point will have the same slope.

The equation is in slope-intercept form, $y = mx + c$, where $m$ is the slope. Therefore, the slope of the given line is $m_{line} = 3$.

Step 2: Calculate the derivative of the curve to find the slope of the tangent.

We are given the curve $y = x^2 + 7x + 2$. We need to find the derivative, $\frac{dy}{dx}$, to determine the slope of the tangent to the curve at any point.

$\frac{dy}{dx} = \frac{d}{dx}(x^2 + 7x + 2)$ $\frac{dy}{dx} = 2x + 7$

The derivative, $2x + 7$, represents the slope of the tangent line to the curve at any $x$-value.

Step 3: Find the x-coordinate of point P(h, k).

The point P on the curve closest to the line will have a tangent with the same slope as the line. So, we set the derivative equal to the slope of the line:

$\frac{dy}{dx} = m_{line}$ $2x + 7 = 3$ Now we solve for $x$: $2x = 3 - 7$ $2x = -4$ $x = -2$ Since P is (h, k), we have $h = -2$.

Step 4: Find the y-coordinate of point P(h, k).

Now that we have $h = -2$, we can find the $y$-coordinate, $k$, by substituting $x = -2$ into the equation of the curve:

$y = x^2 + 7x + 2$ $k = (-2)^2 + 7(-2) + 2$ $k = 4 - 14 + 2$ $k = -8$ Therefore, the point P is $(-2, -8)$.

Step 5: Find the slope of the normal at point P.

The normal to the curve at point P is perpendicular to the tangent at point P. The slope of the tangent at P is $3$ (same as the given line). Therefore, the slope of the normal at P is the negative reciprocal of $3$:

$m_{normal} = -\frac{1}{m_{tangent}} = -\frac{1}{3}$

Step 6: Find the equation of the normal to the curve at point P.

We have the point P $(-2, -8)$ and the slope of the normal $m_{normal} = -\frac{1}{3}$. We can use the point-slope form of a line:

$y - y_1 = m(x - x_1)$ $y - (-8) = -\frac{1}{3}(x - (-2))$ $y + 8 = -\frac{1}{3}(x + 2)$ Multiply both sides by 3: $3(y + 8) = -(x + 2)$ $3y + 24 = -x - 2$ $x + 3y + 26 = 0$

Common Mistakes & Tips

• Remember that the shortest distance is when the tangent is parallel to the line, not perpendicular.
• Be careful with signs when calculating the slope of the normal (negative reciprocal).
• Double-check your arithmetic, especially when substituting values into equations.

Summary

We found the point P on the curve closest to the given line by setting the derivative of the curve equal to the slope of the line. We then found the coordinates of P by substituting the x-value back into the original equation. Finally, we calculated the slope of the normal at P and used the point-slope form to find the equation of the normal, which is $x + 3y + 26 = 0$.

Final Answer

The final answer is \boxed{x + 3y + 26 = 0}, which corresponds to option (D).