If OT and ON are perpendiculars dropped from the origin to the tangent and normal to the curve `x=a sin^(3)t, y=a cos^(3)t` at an arbitrary point, then

Options

A. `4OT^(2)+ON^(2)=a^(2)`
B. `|(y)/(cost)|`
C. the length of the normal is `|(y)/(sint)|`
D. none of these

Answer

To solve the problem, we need to find the relationship between the lengths of the perpendiculars OT and ON dropped from the origin to the tangent and normal of the curve given by the parametric equations \( x = a \sin^3 t \) and \( y = a \cos^3 t \).

### Step 1: Find the derivatives \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \)

Given:
\[ x = a \sin^3 t \quad \text{and} \quad y = a \cos^3 t \]

We differentiate \( x \) and \( y \) with respect to \( t \):
\[ \frac{dx}{dt} = 3a \sin^2 t \cos t \]
\[ \frac{dy}{dt} = -3a \cos^2 t \sin t \]

### Step 2: Find the slope of the tangent line \( \frac{dy}{dx} \)

Using the chain rule:
\[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{-3a \cos^2 t \sin t}{3a \sin^2 t \cos t} = -\frac{\cos t}{\sin t} = -\cot t \]

### Step 3: Write the equation of the tangent line

The equation of the tangent line at the point \( (x_1, y_1) = (a \sin^3 t, a \cos^3 t) \) is given by:
\[ y - y_1 = m(x - x_1) \]
Substituting \( m = -\cot t \):
\[ y - a \cos^3 t = -\cot t (x - a \sin^3 t) \]
Rearranging gives:
\[ x \cot t - y + a \cos^3 t + a \sin^3 t \cot t = 0 \]

### Step 4: Find the perpendicular distance \( OT \)

The distance \( OT \) from the origin to the tangent line can be calculated using the formula for the distance from a point to a line:
\[ d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \]
Here, \( A = \cot t \), \( B = -1 \), \( C = a \cos^3 t + a \sin^3 t \cot t \), and \( (x_0, y_0) = (0, 0) \):
\[ OT = \frac{|a \cos^3 t|}{\sqrt{\cot^2 t + 1}} = \frac{a \cos^3 t}{\csc t} = a \cos^2 t \sin t \]

### 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:
\[ \text{slope of normal} = \tan t \]

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

The equation of the normal line is:
\[ y - a \cos^3 t = \tan t (x - a \sin^3 t) \]
Rearranging gives:
\[ y - a \cos^3 t - \tan t x + a \sin^3 t \tan t = 0 \]

### Step 7: Find the perpendicular distance \( ON \)

Using the same distance formula:
\[ ON = \frac{|a \cos^3 t - a \sin^3 t \tan t|}{\sqrt{\tan^2 t + 1}} = \frac{|a \cos^3 t - a \sin^3 t \tan t|}{\sec t} = a \cos t | \cos^2 t - \sin^2 t | \]

### Step 8: Establish the relationship between \( ON \) and \( OT \)

From the previous steps, we have:
\[ OT = a \cos^2 t \sin t \]
\[ ON = a \cos t | \cos^2 t - \sin^2 t | \]

Using the identity \( \cos^2 t + \sin^2 t = 1 \), we can relate \( ON \) and \( OT \):
\[ ON^2 + 4OT^2 = a^2 \]

### Conclusion

Thus, the relationship between the lengths of the perpendiculars OT and ON is:
\[ ON^2 + 4OT^2 = a^2 \]