Find the equation of normal to the curve `x = a cos^(3)theta, y=b sin^(3) theta" at point "'theta'.`

Answer

To find the equation of the normal to the curve given by the parametric equations \( x = a \cos^3 \theta \) and \( y = b \sin^3 \theta \) at a point defined by the parameter \( \theta \), we can follow these steps:

### Step 1: Find the coordinates of the point on the curve
The coordinates \( (x_1, y_1) \) at the point \( \theta \) are given by:
\[ x_1 = a \cos^3 \theta \]
\[ y_1 = b \sin^3 \theta \]

### Step 2: Differentiate the parametric equations
To find the slope of the tangent line, we need to differentiate \( x \) and \( y \) with respect to \( \theta \):
\[ \frac{dx}{d\theta} = \frac{d}{d\theta}(a \cos^3 \theta) = a \cdot 3 \cos^2 \theta \cdot (-\sin \theta) = -3a \cos^2 \theta \sin \theta \]
\[ \frac{dy}{d\theta} = \frac{d}{d\theta}(b \sin^3 \theta) = b \cdot 3 \sin^2 \theta \cdot \cos \theta = 3b \sin^2 \theta \cos \theta \]

### Step 3: Find the slope of the tangent line
The slope of the tangent line \( \frac{dy}{dx} \) is given by:
\[ \frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{3b \sin^2 \theta \cos \theta}{-3a \cos^2 \theta \sin \theta} \]
This simplifies to:
\[ \frac{dy}{dx} = -\frac{b \sin \theta}{a \cos \theta} = -\frac{b}{a} \tan \theta \]

### Step 4: Find the slope of the normal line
The slope of the normal line is the negative reciprocal of the slope of the tangent line:
\[ \text{slope of normal} = -\frac{1}{\left(-\frac{b}{a} \tan \theta\right)} = \frac{a}{b} \cot \theta \]

### Step 5: Write the equation of the normal line
The equation of the normal line in point-slope form is given by:
\[ y - y_1 = m(x - x_1) \]
Substituting \( y_1 \), \( x_1 \), and the slope of the normal:
\[ y - b \sin^3 \theta = \frac{a}{b} \cot \theta (x - a \cos^3 \theta) \]

### Step 6: Rearranging the equation
Multiplying through by \( b \sin \theta \) to eliminate the fraction:
\[ b \sin \theta (y - b \sin^3 \theta) = a \cos \theta (x - a \cos^3 \theta) \]
Expanding both sides:
\[ b y \sin \theta - b^2 \sin^4 \theta = a x \cos \theta - a^2 \cos^4 \theta \]
Rearranging gives us:
\[ b y \sin \theta - a x \cos \theta = b^2 \sin^4 \theta - a^2 \cos^4 \theta \]

### Final Result
The equation of the normal to the curve at the point defined by \( \theta \) is:
\[ b y \sin \theta - a x \cos \theta = b^2 \sin^4 \theta - a^2 \cos^4 \theta \]