Let `y^(2) -5y +3x +k = 0` be a parabola, then

Options

A. its latus rectum is least when `k = 1`
B. its latus rectum is independent of k
C. the line `y = 2x +1` will touch the parabola if `k = (73)/(16)`
D. `y = (5)/(2)` is the only normal to the parabola whose slope is zero

Explanation

To solve the problem, we need to analyze the given equation of the parabola and verify the options based on the conditions provided. The equation given is:

\[ y^2 - 5y + 3x + k = 0 \]

### Step 1: Rearranging the Equation
We can rearrange the equation to express it in a standard form of a parabola. We will complete the square for the \(y\) terms.

1. **Complete the square for \(y\)**:
\[ y^2 - 5y = (y - \frac{5}{2})^2 - \frac{25}{4} \]
Thus, the equation becomes:
\[ (y - \frac{5}{2})^2 - \frac{25}{4} + 3x + k = 0 \]
Rearranging gives:
\[ (y - \frac{5}{2})^2 = -3x + \frac{25}{4} - k \]

### Step 2: Identify the Form of the Parabola
The equation can be rewritten as:
\[ (y - \frac{5}{2})^2 = -3(x - \frac{25 - 4k}{12}) \]
This is in the form \((y - k)^2 = 4a(x - h)\), where:
- \(4a = -3\)
- The parabola opens to the left since \(4a < 0\).

### Step 3: Length of the Latus Rectum
The length of the latus rectum \(L\) is given by \(L = |4a|\). Here, since \(4a = -3\):
\[ L = 3 \]
This length is independent of \(k\).

### Step 4: Finding the Condition for Tangency
Next, we need to check if a line can touch the parabola. Let's assume the line has the equation:
\[ y = mx + c \]
Substituting \(y\) into the parabola's equation:
\[ (mx + c)^2 - 5(mx + c) + 3x + k = 0 \]
Expanding this gives:
\[ m^2x^2 + 2mcx + c^2 - 5mx - 5c + 3x + k = 0 \]
This is a quadratic in \(x\). For the line to be tangent to the parabola, the discriminant must be zero:
\[ D = (2mc - 5m + 3)^2 - 4m^2(c^2 - 5c + k) = 0 \]

### Step 5: Solving for \(k\)
Solving the discriminant condition will yield a specific value for \(k\). After simplification, we find:
\[ k = \frac{73}{16} \]

### Step 6: Normal to the Parabola
To find the normal to the parabola, we first find the slope of the tangent:
\[ \frac{dy}{dx} = -\frac{3}{2}(y - \frac{5}{2}) \]
The slope of the normal is the negative reciprocal:
\[ \text{slope of normal} = \frac{2(y - \frac{5}{2})}{3} \]
Setting this equal to zero gives:
\[ y - \frac{5}{2} = 0 \implies y = \frac{5}{2} \]
This indicates that the normal at this point is horizontal.

### Conclusion
From the analysis, we conclude that:
1. The length of the latus rectum is independent of \(k\).
2. The specific value of \(k\) for tangency is \(k = \frac{73}{16}\).
3. The normal at \(y = \frac{5}{2}\) is valid.

Thus, the correct options are the second, third, and fourth.