The equation, `2x^2+ 3y^2-8x-18y+35= K` represents (a) no locus if ` k gt 0` (b) an ellipse if ` k lt 0` (c) a point if ` k = 0` (d) a hyperbola if `k gt0`

Options

A. no locus if k gt 0
B. an ellipse if k lt 0
C. a point if k = 0
D. a hyperbola if k gt0

Explanation

To solve the problem, we need to analyze the equation \(2x^2 + 3y^2 - 8x - 18y + 35 = K\) and determine what it represents for different values of \(K\).

### Step 1: Rearranging the equation
We start with the equation:
\[ 2x^2 + 3y^2 - 8x - 18y + 35 = K \]
We can rearrange it as:
\[ 2x^2 - 8x + 3y^2 - 18y + 35 - K = 0 \]

### Step 2: Completing the square for \(x\) and \(y\)
Next, we complete the square for the \(x\) and \(y\) terms.

For \(x\):
\[ 2(x^2 - 4x) = 2((x - 2)^2 - 4) = 2(x - 2)^2 - 8 \]

For \(y\):
\[ 3(y^2 - 6y) = 3((y - 3)^2 - 9) = 3(y - 3)^2 - 27 \]

Substituting these back into the equation gives:
\[ 2(x - 2)^2 - 8 + 3(y - 3)^2 - 27 + 35 = K \]
Simplifying this, we have:
\[ 2(x - 2)^2 + 3(y - 3)^2 = K + 8 + 27 - 35 \]
\[ 2(x - 2)^2 + 3(y - 3)^2 = K \]

### Step 3: Analyzing the equation
Now, we analyze the equation \(2(x - 2)^2 + 3(y - 3)^2 = K\).

1. **If \(K > 0\)**: The left side \(2(x - 2)^2 + 3(y - 3)^2\) is always non-negative (since squares are non-negative). Thus, there are real solutions, and the equation represents an ellipse.

2. **If \(K = 0\)**: The equation becomes \(2(x - 2)^2 + 3(y - 3)^2 = 0\). This is only true when both squares are zero, which means:
\[ x - 2 = 0 \quad \text{and} \quad y - 3 = 0 \]
Therefore, \(x = 2\) and \(y = 3\), which represents a single point.

3. **If \(K < 0\)**: The left side \(2(x - 2)^2 + 3(y - 3)^2\) cannot be negative, so there are no real solutions. Thus, the equation represents no locus.

### Conclusion
Based on the analysis:
- \(K > 0\) represents an ellipse.
- \(K = 0\) represents a point at \((2, 3)\).
- \(K < 0\) represents no locus.

### Final Answer
The correct options are:
- (a) no locus if \(K < 0\)
- (b) an ellipse if \(K > 0\)
- (c) a point if \(K = 0\)