For the variable t, the locus of the point of intersection of the lines `3tx-2y+6t=0 and 3x+2ty-6=0` is

Options

A. the ellipse `(x^(2))/(4)+(y^(2))/(9)=1`
B. the ellipse `(x^(2))/(9)+(y^(2))/(4)=1`
C. The hyperbola `(x^(2))/(4)-(y^(2))/(9)=1`
D. the hyperbola `(x^(2))/(9)-(y^(2))/(4)=1`

Answer

To find the locus of the point of intersection of the lines given by the equations \(3tx - 2y + 6t = 0\) and \(3x + 2ty - 6 = 0\), we will follow these steps:

### Step 1: Solve for the intersection point
We have two equations:
1. \(3tx - 2y + 6t = 0\) (Equation 1)
2. \(3x + 2ty - 6 = 0\) (Equation 2)

We can express \(y\) in terms of \(x\) from Equation 1:
\[ 2y = 3tx + 6t \implies y = \frac{3tx + 6t}{2} \]

### Step 2: Substitute \(y\) into Equation 2
Now substitute \(y\) from Equation 1 into Equation 2:
\[ 3x + 2t\left(\frac{3tx + 6t}{2}\right) - 6 = 0 \]
This simplifies to:
\[ 3x + 3tx + 6t - 6 = 0 \]
Combining like terms gives:
\[ (3 + 3t)x + 6t - 6 = 0 \]

### Step 3: Solve for \(x\)
Rearranging the equation:
\[ (3 + 3t)x = 6 - 6t \implies x = \frac{6 - 6t}{3 + 3t} \]
This can be simplified to:
\[ x = \frac{2(1 - t)}{1 + t} \]

### Step 4: Substitute \(x\) back to find \(y\)
Now substitute \(x\) back into the expression for \(y\):
\[ y = \frac{3t\left(\frac{2(1 - t)}{1 + t}\right) + 6t}{2} \]
Simplifying this gives:
\[ y = \frac{3t \cdot \frac{2(1 - t)}{1 + t} + 6t}{2} \]
\[ = \frac{6t(1 - t) + 6t(1 + t)}{2(1 + t)} \]
\[ = \frac{6t}{2(1 + t)} = \frac{3t}{1 + t} \]

### Step 5: Eliminate the parameter \(t\)
Now we have \(x\) and \(y\) in terms of \(t\):
\[ x = \frac{2(1 - t)}{1 + t} \quad \text{and} \quad y = \frac{3t}{1 + t} \]
We can express \(t\) in terms of \(x\) from the equation for \(x\):
\[ x(1 + t) = 2(1 - t) \implies x + xt = 2 - 2t \implies (x + 2)t = 2 - x \implies t = \frac{2 - x}{x + 2} \]

Substituting \(t\) into the equation for \(y\):
\[ y = \frac{3\left(\frac{2 - x}{x + 2}\right)}{1 + \frac{2 - x}{x + 2}} = \frac{3(2 - x)}{(x + 2) + (2 - x)} = \frac{3(2 - x)}{4} \]

### Step 6: Find the locus equation
Now we can eliminate \(t\) and find the relationship between \(x\) and \(y\):
\[ y = \frac{3(2 - x)}{4} \implies 4y = 6 - 3x \implies 3x + 4y - 6 = 0 \]

### Step 7: Identify the type of locus
To find the locus as a conic section, we can rearrange the equation:
\[ 3x + 4y = 6 \]
This is a linear equation, indicating that the locus is a straight line.

### Final Answer
The locus of the point of intersection of the lines is given by the equation:
\[ 3x + 4y - 6 = 0 \]

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