Equation of tangent to the curve `x = a cos^3 t, y=asin^3t` at `'t'` is

Options

A. x sec t-y cosec t =a
B. x sec t + y cosec t =a
C. `xcosect+ycosect=a`
D. None of these

Answer

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

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

### Step 2: Compute \( \frac{dy}{dt} \) and \( \frac{dx}{dt} \)
To find the slope of the tangent line, we need to compute the derivatives of \( x \) and \( y \) with respect to \( t \).

For \( y \):
\[ \frac{dy}{dt} = \frac{d}{dt}(a \sin^3 t) = 3a \sin^2 t \cdot \cos t \]

For \( x \):
\[ \frac{dx}{dt} = \frac{d}{dt}(a \cos^3 t) = -3a \cos^2 t \cdot \sin t \]

### Step 3: Find \( \frac{dy}{dx} \)
The slope \( m \) of the tangent line is given by:
\[ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{3a \sin^2 t \cdot \cos t}{-3a \cos^2 t \cdot \sin t} \]
Simplifying this, we get:
\[ \frac{dy}{dx} = -\frac{\sin t}{\cos t} = -\tan t \]

### Step 4: Write the equation of the tangent line
Using the point-slope form of the equation of a line, \( y - y_1 = m(x - x_1) \):
\[ y - a \sin^3 t = -\tan t (x - a \cos^3 t) \]

### Step 5: Rearranging the equation
Expanding and rearranging gives:
\[ y - a \sin^3 t = -\tan t \cdot x + a \tan t \cdot \cos^3 t \]
Rearranging terms:
\[ y + x \tan t = a \tan t \cdot \cos^3 t + a \sin^3 t \]

### Step 6: Final form of the equation
We can express this in a more standard form:
\[ x \tan t + y = a (\tan t \cdot \cos^3 t + \sin^3 t) \]