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

Options

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

Answer

To find the equation of the tangent 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 \( (x_1, y_1) \) at \( \theta = \frac{\pi}{4} \)
We substitute \( \theta = \frac{\pi}{4} \) into the equations for \( x \) and \( y \):

\[ x_1 = a \cos^3 \left( \frac{\pi}{4} \right) = a \left( \frac{1}{\sqrt{2}} \right)^3 = \frac{a}{2\sqrt{2}} \]

\[ y_1 = a \sin^3 \left( \frac{\pi}{4} \right) = a \left( \frac{1}{\sqrt{2}} \right)^3 = \frac{a}{2\sqrt{2}} \]

### Step 2: Calculate \( \frac{dx}{d\theta} \) and \( \frac{dy}{d\theta} \)
Next, we find the derivatives \( \frac{dx}{d\theta} \) and \( \frac{dy}{d\theta} \):

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

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

### Step 3: Find \( \frac{dy}{dx} \)
Using the chain rule, we find \( \frac{dy}{dx} \):

\[ \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{3a \sin^2 \theta \cos \theta}{-3a \cos^2 \theta \sin \theta} \]

Simplifying this gives:

\[ \frac{dy}{dx} = -\frac{\sin \theta}{\cos \theta} = -\tan \theta \]

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

\[ \frac{dy}{dx} \bigg|_{\theta = \frac{\pi}{4}} = -\tan \left( \frac{\pi}{4} \right) = -1 \]

### Step 5: Write the equation of the tangent line
Using the point-slope form of the equation of a line, we have:

\[ y - y_1 = m(x - x_1) \]

Where \( m = -1 \), \( x_1 = \frac{a}{2\sqrt{2}} \), and \( y_1 = \frac{a}{2\sqrt{2}} \):

\[ y - \frac{a}{2\sqrt{2}} = -1 \left( x - \frac{a}{2\sqrt{2}} \right) \]

### Step 6: Simplify the equation
Rearranging gives:

\[ y - \frac{a}{2\sqrt{2}} = -x + \frac{a}{2\sqrt{2}} \]

\[ y + x = \frac{a}{\sqrt{2}} \]

### Final Equation
Multiplying through by \( \sqrt{2} \):

\[ \sqrt{2}x + \sqrt{2}y - a = 0 \]

Thus, the equation of the tangent line is:

\[ \sqrt{2}x + \sqrt{2}y - a = 0 \]