Key Concepts and Formulas

• Parametric Differentiation: Given $x = f(\theta)$ and $y = g(\theta)$, then $\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}$.
• Slope of Tangent and Normal: If the slope of the tangent is $m_T$, then the slope of the normal, $m_N$, is given by $m_N = -\frac{1}{m_T}$.
• Equation of a Line: The equation of a line with slope $m$ passing through point $(x_1, y_1)$ is $y - y_1 = m(x - x_1)$.

Step-by-Step Solution

Step 1: Find $\frac{dx}{d\theta}$ and $\frac{dy}{d\theta}$

We are given $x = a(\cos \theta + \theta \sin \theta)$ and $y = a(\sin \theta - \theta \cos \theta)$. We need to find the derivatives of $x$ and $y$ with respect to $\theta$. This will allow us to calculate the slope of the tangent.

$\frac{dx}{d\theta} = a(-\sin \theta + \sin \theta + \theta \cos \theta) = a\theta \cos \theta$ $\frac{dy}{d\theta} = a(\cos \theta - \cos \theta + \theta \sin \theta) = a\theta \sin \theta$

Step 2: Find $\frac{dy}{dx}$ (slope of the tangent)

Now, we find the slope of the tangent by dividing $\frac{dy}{d\theta}$ by $\frac{dx}{d\theta}$: $\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{a\theta \sin \theta}{a\theta \cos \theta} = \tan \theta$

Step 3: Find the slope of the normal

The slope of the normal is the negative reciprocal of the slope of the tangent: $m_N = -\frac{1}{\tan \theta} = -\cot \theta$

Step 4: Find the equation of the normal at $\theta = \theta'$

The equation of the normal at the point where $\theta = \theta'$ is given by: $y - a(\sin \theta' - \theta' \cos \theta') = -\cot \theta' [x - a(\cos \theta' + \theta' \sin \theta')]$ $y - a\sin \theta' + a\theta' \cos \theta' = -\frac{\cos \theta'}{\sin \theta'} [x - a\cos \theta' - a\theta' \sin \theta']$ $y\sin \theta' - a\sin^2 \theta' + a\theta' \cos \theta' \sin \theta' = -x\cos \theta' + a\cos^2 \theta' + a\theta' \sin \theta' \cos \theta'$ $x\cos \theta' + y\sin \theta' = a\cos^2 \theta' + a\sin^2 \theta'$ $x\cos \theta' + y\sin \theta' = a(\cos^2 \theta' + \sin^2 \theta')$ $x\cos \theta' + y\sin \theta' = a$

Step 5: Check if the normal passes through the origin

If the normal passes through the origin $(0, 0)$, then substituting $x = 0$ and $y = 0$ into the equation of the normal should satisfy the equation: $(0)\cos \theta' + (0)\sin \theta' = a$ $0 = a$ However, $a$ is a constant (presumably non-zero). This seems to suggest that the normal does not pass through the origin. However, if we look at the equation of the normal, $x\cos \theta' + y\sin \theta' = a$, we can calculate the distance from the origin to the normal.

Step 6: Find the distance of the normal from the origin.

The distance of the line $Ax + By + C = 0$ from the origin is given by $\frac{|C|}{\sqrt{A^2 + B^2}}$. Rewriting our normal equation, we have $x\cos \theta' + y\sin \theta' - a = 0$. Thus, the distance is $d = \frac{|-a|}{\sqrt{\cos^2 \theta' + \sin^2 \theta'}} = \frac{a}{\sqrt{1}} = a$ This is a constant distance.

Common Mistakes & Tips

• Carefully apply the chain rule when finding $\frac{dx}{d\theta}$ and $\frac{dy}{d\theta}$.
• Remember that the slope of the normal is the negative reciprocal of the slope of the tangent.
• The distance from a point $(x_0, y_0)$ to the line $Ax + By + C = 0$ is given by $\frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$. When finding the distance from the origin, $(x_0, y_0) = (0,0)$.

Summary

We found the derivatives $\frac{dx}{d\theta}$ and $\frac{dy}{d\theta}$, then calculated the slope of the tangent and the normal. We then found the equation of the normal and determined that it is at a constant distance $a$ from the origin. This corresponds to option (D). However, the given answer is (A). Let's review the question and our working.

If the normal passes through the origin, then the distance of the normal to the origin is zero. We calculated the distance to be $a$. Therefore, the normal does not pass through the origin unless $a=0$ which is not possible.

The problem states that the correct answer is (A). Let's re-examine the equation of the normal: $x\cos \theta' + y\sin \theta' = a$. This equation is equivalent to: $\frac{x}{a/\cos\theta'} + \frac{y}{a/\sin\theta'} = 1$ Let's reconsider the question statement and the given answer. The equation $x\cos \theta' + y\sin \theta' = a$ can be rewritten as:

$x \cos \theta' + y \sin \theta' = a$

The distance from the origin to this line is $\frac{|a|}{\sqrt{\cos^2 \theta' + \sin^2 \theta'}} = a$, which is constant. Therefore, the normal is at a constant distance from the origin.

If the normal passed through the origin, substituting $x=0$ and $y=0$ into $x \cos \theta' + y \sin \theta' = a$ would yield $0 = a$, which is generally false.

Therefore, the correct answer must be that the normal is at a constant distance from the origin, corresponding to option (D).

Final Answer

The final answer is \boxed{D}, which corresponds to option (D).