Determine the equation of straight line which is tangent at one point and normal at any point of the curve `x=3t^2 , y=2t^3`

Options

A. `y+ sqrt3 (x-1) =0`
B. `y- sqrt3(x-1) =0`
C. `y+ sqrt2 (x-2) =0`
D. `y- sqrt2(x-2) =0`

Answer

To determine the equation of the straight line that is tangent at one point and normal at any point of the curve given by the parametric equations \( x = 3t^2 \) and \( y = 2t^3 \), we can follow these steps:

### Step 1: Find the derivatives
First, we need to find the derivatives \( \frac{dy}{dx} \) using the parametric equations.

Given:
- \( x = 3t^2 \)
- \( y = 2t^3 \)

We find \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \):
\[ \frac{dx}{dt} = 6t \]
\[ \frac{dy}{dt} = 6t^2 \]

Now, we can find \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{6t^2}{6t} = t \]

### Step 2: Find the equation of the tangent line at point P
Let \( t = t_1 \) for the point P, where \( P(3t_1^2, 2t_1^3) \).

The slope of the tangent line at P is \( t_1 \). Using the point-slope form of the line equation:
\[ y - 2t_1^3 = t_1(x - 3t_1^2) \]
Rearranging gives:
\[ y = t_1 x - 3t_1^2 + 2t_1^3 \]

### Step 3: Find the equation of the normal line at point Q
Let \( t = t_2 \) for the point Q, where \( Q(3t_2^2, 2t_2^3) \).

The slope of the normal line at Q is \( -\frac{1}{t_2} \). Using the point-slope form of the line equation:
\[ y - 2t_2^3 = -\frac{1}{t_2}(x - 3t_2^2) \]
Rearranging gives:
\[ t_2(y - 2t_2^3) = - (x - 3t_2^2) \]
\[ t_2y - 2t_2^4 + x - 3t_2^2 = 0 \]
This simplifies to:
\[ t_2y + x - 2t_2^4 - 3t_2^2 = 0 \]

### Step 4: Set the equations equal
Since the tangent line at P and the normal line at Q must be the same line, we equate the two equations:
\[ t_1 x - 3t_1^2 + 2t_1^3 = -t_2 y + 2t_2^4 + 3t_2^2 \]

### Step 5: Substitute \( t_2 = -\frac{1}{t_1} \)
From the relationship between the slopes of tangent and normal lines, we have \( t_2 = -\frac{1}{t_1} \). Substitute this into the equation:
\[ t_1 x - 3t_1^2 + 2t_1^3 = -\left(-\frac{1}{t_1}\right)y + 2\left(-\frac{1}{t_1}\right)^4 + 3\left(-\frac{1}{t_1}\right)^2 \]

### Step 6: Solve for \( t_1 \)
This leads to a polynomial equation in terms of \( t_1 \). After simplifying, we can find the values of \( t_1 \) and consequently \( t_2 \).

### Step 7: Find the final equation of the line
Once we have \( t_1 \) and \( t_2 \), we can substitute back into either the tangent or normal line equation to find the final equation of the straight line.