The equations of two sides of a variable triangle are `x=0` and `y=3`,and its third side is a tangent to the parabola `y^(2)=6x`.The locus of its circumcentre is:

Options

A. `4y^(2)-18y+3x+18=0`
B. `4y^(2)-18y-3x+18=0` (3)
C. `4y^(2)+18y+3x+18=0`
D. `4y^(2)-18y-3x-18=0`

Answer

To solve the problem, we need to find the locus of the circumcenter of a triangle formed by the lines \(x = 0\), \(y = 3\), and a tangent to the parabola \(y^2 = 6x\).

### Step-by-step Solution:

1. **Identify the Fixed Points**:
The lines \(x = 0\) and \(y = 3\) intersect at the point \(A(0, 3)\). The line \(x = 0\) is the y-axis, and \(y = 3\) is a horizontal line.

2. **Equation of the Parabola**:
The equation of the parabola is given by \(y^2 = 6x\). We can rewrite this in standard form as \(x = \frac{y^2}{6}\).

3. **Finding the Tangent to the Parabola**:
The equation of the tangent to the parabola \(y^2 = 6x\) at a point \((x_0, y_0)\) can be expressed as:
\[ yy_0 = 3(x + x_0) \]
For a point on the parabola, we can express \(x_0\) in terms of \(y_0\):
\[ x_0 = \frac{y_0^2}{6} \]
Substituting this into the tangent equation gives:
\[ yy_0 = 3\left(x + \frac{y_0^2}{6}\right) \]
Rearranging this, we get:
\[ yy_0 = 3x + \frac{y_0^2}{2} \]

4. **Finding the Coordinates of the Third Vertex**:
The third vertex \(C\) of the triangle lies on the tangent line. Let’s denote the point on the tangent line as \(C(x_C, y_C)\). Since the tangent line intersects the line \(y = 3\), we can set \(y_C = 3\).

5. **Finding the Circumcenter**:
The circumcenter \(O\) of triangle \(ABC\) can be found as the midpoint of the line segment joining points \(A(0, 3)\) and \(C(x_C, 3)\). The coordinates of the circumcenter \(O(h, k)\) are:
\[ h = \frac{0 + x_C}{2} = \frac{x_C}{2}, \quad k = \frac{3 + 3}{2} = 3 \]

6. **Expressing \(x_C\) in terms of \(h\)**:
From the expression for \(h\), we have:
\[ x_C = 2h \]

7. **Substituting \(x_C\) into the Tangent Equation**:
Substitute \(y_C = 3\) into the tangent equation:
\[ 3y_0 = 3\left(2h + \frac{y_0^2}{6}\right) \]
Simplifying this gives:
\[ 3y_0 = 6h + \frac{y_0^2}{2} \]
Rearranging leads to:
\[ y_0^2 - 6y_0 + 12h = 0 \]

8. **Finding the Locus**:
The discriminant of this quadratic must be non-negative for real \(y_0\):
\[ (-6)^2 - 4 \cdot 1 \cdot (12h) \geq 0 \]
Simplifying gives:
\[ 36 - 48h \geq 0 \implies h \leq \frac{3}{4} \]
Thus, the locus of the circumcenter is:
\[ h = \frac{x}{2}, \quad k = 3 \]
This describes a vertical line at \(h = \frac{3}{4}\).

### Final Locus Equation:
The locus of the circumcenter is given by:
\[ x = \frac{3}{4} \]