The equation of normal to the curve `x= a cos^(3) theta, y=a sin^(3) theta" at "theta=(pi)/(4)` is

Options

A. x-y=0
B. x+y=0
C. `x-y=asqrt(2)`
D. `x+y=asqrt(2)`

Answer

To find the equation of the normal to the curve given by \( x = a \cos^3 \theta \) and \( y = a \sin^3 \theta \) at \( \theta = \frac{\pi}{4} \), we will follow these steps:

### Step 1: Find the coordinates of the point on the curve at \( \theta = \frac{\pi}{4} \).

Using the parametric equations:
\[ x = a \cos^3 \theta \]
\[ y = a \sin^3 \theta \]

Substituting \( \theta = \frac{\pi}{4} \):
\[ x_1 = a \left(\cos \frac{\pi}{4}\right)^3 = a \left(\frac{1}{\sqrt{2}}\right)^3 = a \cdot \frac{1}{2\sqrt{2}} = \frac{a}{2\sqrt{2}} \]
\[ y_1 = a \left(\sin \frac{\pi}{4}\right)^3 = a \left(\frac{1}{\sqrt{2}}\right)^3 = a \cdot \frac{1}{2\sqrt{2}} = \frac{a}{2\sqrt{2}} \]

### Step 2: Differentiate \( x \) and \( y \) with respect to \( \theta \).

Calculating \( \frac{dx}{d\theta} \):
\[ \frac{dx}{d\theta} = a \cdot 3 \cos^2 \theta \cdot (-\sin \theta) = -3a \cos^2 \theta \sin \theta \]

Calculating \( \frac{dy}{d\theta} \):
\[ \frac{dy}{d\theta} = a \cdot 3 \sin^2 \theta \cdot \cos \theta = 3a \sin^2 \theta \cos \theta \]

### Step 3: Find \( \frac{dy}{dx} \).

Using the chain rule:
\[ \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{3a \sin^2 \theta \cos \theta}{-3a \cos^2 \theta \sin \theta} \]
This simplifies to:
\[ \frac{dy}{dx} = -\frac{\sin \theta}{\cos \theta} = -\tan \theta \]

### Step 4: Evaluate \( \frac{dy}{dx} \) at \( \theta = \frac{\pi}{4} \).

Substituting \( \theta = \frac{\pi}{4} \):
\[ \frac{dy}{dx} = -\tan \frac{\pi}{4} = -1 \]

### Step 5: Find the slope of the normal line.

The slope of the normal line is the negative reciprocal of the slope of the tangent line:
\[ \text{slope of normal} = -\frac{1}{-1} = 1 \]

### Step 6: Write the equation of the normal line.

Using the point-slope form of the equation of a line:
\[ y - y_1 = m(x - x_1) \]
Substituting \( m = 1 \), \( x_1 = \frac{a}{2\sqrt{2}} \), and \( y_1 = \frac{a}{2\sqrt{2}} \):
\[ y - \frac{a}{2\sqrt{2}} = 1 \left( x - \frac{a}{2\sqrt{2}} \right) \]

This simplifies to:
\[ y - \frac{a}{2\sqrt{2}} = x - \frac{a}{2\sqrt{2}} \]
Rearranging gives:
\[ x - y = 0 \]

### Final Answer:
The equation of the normal to the curve at \( \theta = \frac{\pi}{4} \) is:
\[ x - y = 0 \]