Key Concepts and Formulas

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

Step-by-Step Solution

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

We are given $x = a(1 + \cos \theta)$ and $y = a\sin \theta$. We need to find the derivatives of $x$ and $y$ with respect to $\theta$.

$\frac{dx}{d\theta} = \frac{d}{d\theta} [a(1 + \cos \theta)] = a(0 - \sin \theta) = -a\sin \theta$ $\frac{dy}{d\theta} = \frac{d}{d\theta} [a\sin \theta] = a\cos \theta$

Step 2: Find $\frac{dy}{dx}$

Using the formula for parametric differentiation, we have:

$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{a\cos \theta}{-a\sin \theta} = -\cot \theta$

This is the slope of the tangent to the curve at the point corresponding to $\theta$.

Step 3: Find the slope of the normal

The slope of the normal is the negative reciprocal of the slope of the tangent. Therefore, the slope of the normal is:

$m_{normal} = -\frac{1}{dy/dx} = -\frac{1}{-\cot \theta} = \tan \theta$

Step 4: Find the equation of the normal

The equation of the normal to the curve at the point $(x, y) = (a(1 + \cos \theta), a\sin \theta)$ with slope $\tan \theta$ is given by:

$y - a\sin \theta = \tan \theta (x - a(1 + \cos \theta))$

Step 5: Simplify the equation of the normal

We can rewrite the equation as:

$y - a\sin \theta = \frac{\sin \theta}{\cos \theta} (x - a - a\cos \theta)$ Multiplying both sides by $\cos \theta$, we get: $y\cos \theta - a\sin \theta \cos \theta = x\sin \theta - a\sin \theta - a\sin \theta \cos \theta$ $y\cos \theta = x\sin \theta - a\sin \theta + a\sin \theta \cos \theta + a\sin \theta \cos \theta$

$y\cos \theta - x\sin \theta = -a\sin \theta$ $x\sin \theta - y\cos \theta = a\sin \theta$

Step 6: Find the fixed point

The equation of the normal is $x\sin \theta - y\cos \theta = a\sin \theta$. We want to find a point $(x_0, y_0)$ that satisfies this equation for all values of $\theta$. Let's test the given options.

• (A) (a, a): Substituting $x = a$ and $y = a$ into the equation gives: $a\sin \theta - a\cos \theta = a\sin \theta$, which simplifies to $-a\cos \theta = 0$. This is only true for specific values of $\theta$, not all values. So, (A) is incorrect.

• (B) (0, a): Substituting $x = 0$ and $y = a$ into the equation gives: $0\cdot\sin \theta - a\cos \theta = a\sin \theta$, which simplifies to $-a\cos \theta = a\sin \theta$ or $\tan \theta = -1$. This is only true for specific values of $\theta$, not all values. So, (B) is incorrect.

• (C) (0, 0): Substituting $x = 0$ and $y = 0$ into the equation gives: $0\cdot\sin \theta - 0\cdot\cos \theta = a\sin \theta$, which simplifies to $0 = a\sin \theta$. This is only true for specific values of $\theta$, not all values. So, (C) is incorrect.

• (D) (a, 0): Substituting $x = a$ and $y = 0$ into the equation gives: $a\sin \theta - 0\cdot\cos \theta = a\sin \theta$, which simplifies to $a\sin \theta = a\sin \theta$. This is true for all values of $\theta$.

Therefore, the normal always passes through the point $(a, 0)$.

3. Common Mistakes & Tips

• Sign Errors: Be careful with the signs when differentiating and finding the negative reciprocal for the slope of the normal.
• Simplification: Simplifying the equation of the normal is crucial for identifying the fixed point.
• Checking the Options: When looking for a fixed point, substitute the coordinates of the potential fixed point into the equation of the normal and see if the equation holds true for all values of the parameter $\theta$.

4. Summary

We found the equation of the normal to the parametric curve $x = a(1 + \cos \theta)$, $y = a\sin \theta$ at the point corresponding to the parameter $\theta$. By substituting the coordinates of the given options into the equation of the normal, we determined that the point $(a, 0)$ satisfies the equation for all values of $\theta$.

5. Final Answer

The final answer is \boxed{(a, 0)}, which corresponds to option (D).