If the tangent at a point `P` with parameter `t`, on the curve `x=4t^2+3`, `y=8t^3-1` `t in R` meets the curve again at a point Q, then the coordinates of Q are

Options

A. `((35)/(9), pm (16sqrt(2))/(27)-1) `
B. `((25)/(9), bot (11)/(7))`
C. `((35)/(9), pm (16sqrt(2))/(27)+1)`
D. none of these

Explanation

To find the coordinates of point Q where the tangent at point P with parameter t meets the curve again, we will follow these steps:

### Step 1: Define the curve
The given parametric equations of the curve are:
- \( x = 4t^2 + 3 \)
- \( y = 8t^3 - 1 \)

### Step 2: Find the derivatives
To find the slope of the tangent, we need to compute \( \frac{dy}{dx} \):
1. Find \( \frac{dx}{dt} \):
\[ \frac{dx}{dt} = \frac{d}{dt}(4t^2 + 3) = 8t \]
2. Find \( \frac{dy}{dt} \):
\[ \frac{dy}{dt} = \frac{d}{dt}(8t^3 - 1) = 24t^2 \]
3. Now, calculate \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{24t^2}{8t} = 3t \]

### Step 3: Write the equation of the tangent line at point P
Using the point-slope form of the equation of a line, the equation of the tangent at point P (where \( t \) is the parameter) is:
\[ y - (8t^3 - 1) = 3t(x - (4t^2 + 3)) \]
Simplifying this gives:
\[ y - 8t^3 + 1 = 3t(x - 4t^2 - 3) \]
Rearranging:
\[ y = 3tx - 12t^3 + 8t^3 + 1 \]
\[ y = 3tx - 4t^3 + 1 \]

### Step 4: Substitute the parametric equations into the tangent line equation
To find where this tangent line meets the curve again, substitute \( x = 4t_1^2 + 3 \) and \( y = 8t_1^3 - 1 \) into the tangent line equation:
\[ 8t_1^3 - 1 = 3t(4t_1^2 + 3) - 4t^3 + 1 \]
This simplifies to:
\[ 8t_1^3 - 1 = 12tt_1^2 + 9t - 4t^3 + 1 \]
Rearranging gives:
\[ 8t_1^3 - 12tt_1^2 - 4t^3 - 9t + 2 = 0 \]

### Step 5: Solve the equation for \( t_1 \)
This is a cubic equation in \( t_1 \). We can factor or use the quadratic formula to find the roots. The equation can be rearranged as:
\[ 2t_1^3 - 6tt_1^2 - 2t^3 - \frac{9}{4}t + 1 = 0 \]
Using the factorization method or synthetic division, we can find the values of \( t_1 \).

### Step 6: Find the coordinates of Q
Once we find \( t_1 \), we can substitute back into the parametric equations to find the coordinates of Q:
1. For \( x \):
\[ x = 4t_1^2 + 3 \]
2. For \( y \):
\[ y = 8t_1^3 - 1 \]

### Step 7: Final coordinates
After solving for \( t_1 \) and substituting back into the equations, we will obtain the coordinates of point Q.