The equation of tangent to the curve ` x=a cos^(3)t ,y=a sin^(3) t ` at 't' is

Options

A. `x sec t- y "cosec" t=a`
B. `x sec t+ y "cosec" t=a`
C. `x "cosec"t+y cos t=a`
D. `x sec t+ y cos t =-a `

Answer

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

### Step 1: Differentiate \( x \) and \( y \) with respect to \( t \)
We start by finding the derivatives of \( x \) and \( y \) with respect to \( t \).

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

\[ y = a \sin^3 t \implies \frac{dy}{dt} = a \cdot 3 \sin^2 t \cdot \cos t = 3a \sin^2 t \cos t \]

### Step 2: Find the slope \( \frac{dy}{dx} \)
Using the chain rule, we can find the slope of the tangent line:

\[ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{3a \sin^2 t \cos t}{-3a \cos^2 t \sin t} \]

Simplifying this expression:

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

### Step 3: Find the point on the curve at parameter \( t \)
The coordinates of the point on the curve at parameter \( t \) are:

\[ (x_1, y_1) = (a \cos^3 t, a \sin^3 t) \]

### Step 4: Write the equation of the tangent line
Using the point-slope form of the equation of a line, we have:

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

Substituting \( y_1 = a \sin^3 t \), \( m = -\tan t \), \( x_1 = a \cos^3 t \):

\[ y - a \sin^3 t = -\tan t (x - a \cos^3 t) \]

### Step 5: Rearranging the equation
Rearranging the equation gives:

\[ y - a \sin^3 t = -\tan t \cdot x + a \tan t \cos^3 t \]

### Step 6: Simplifying the equation
We can multiply through by \( \cos t \) to eliminate the tangent:

\[ y \cos t - a \sin^3 t \cos t = -x + a \cos^3 t \sin t \]

Rearranging gives:

\[ x \sin t + y \cos t = a \]

### Final Equation
Thus, the equation of the tangent to the curve at the point corresponding to \( t \) is:

\[ x \sin t + y \cos t = a \]