The locus of a point which is equidistant from point (4,2) and x-axis is

Options

A. `h^(2)-8h-4k+20=0`
B. `h^(2)-8h+4k-20=0`
C. `h^(2)-6h+4k+20=0`
D. None of these

Answer

To find the locus of a point that is equidistant from the point (4, 2) and the x-axis, we can follow these steps:

### Step 1: Define the point on the locus
Let the point on the locus be denoted as \( (h, k) \).

### Step 2: Calculate the distance from the point to (4, 2)
Using the distance formula, the distance \( d_1 \) from the point \( (h, k) \) to the point \( (4, 2) \) is given by:
\[ d_1 = \sqrt{(h - 4)^2 + (k - 2)^2} \]

### Step 3: Calculate the distance from the point to the x-axis
The distance \( d_2 \) from the point \( (h, k) \) to the x-axis is simply the y-coordinate, which is \( |k| \). Since we are looking for points above or below the x-axis, we can consider it as \( k \) (assuming \( k \geq 0 \)).

### Step 4: Set the distances equal
Since the point is equidistant from (4, 2) and the x-axis, we set these distances equal:
\[ \sqrt{(h - 4)^2 + (k - 2)^2} = k \]

### Step 5: Square both sides to eliminate the square root
Squaring both sides gives:
\[ (h - 4)^2 + (k - 2)^2 = k^2 \]

### Step 6: Expand both sides
Expanding the left-hand side:
\[ (h - 4)^2 = h^2 - 8h + 16 \]
\[ (k - 2)^2 = k^2 - 4k + 4 \]
So, we have:
\[ h^2 - 8h + 16 + k^2 - 4k + 4 = k^2 \]

### Step 7: Simplify the equation
Now, we can simplify the equation:
\[ h^2 - 8h + 20 - 4k = 0 \]
This can be rearranged to:
\[ h^2 - 8h + 20 = 4k \]

### Step 8: Rearranging for k
To express \( k \) in terms of \( h \):
\[ k = \frac{h^2 - 8h + 20}{4} \]

### Step 9: Identify the locus
The equation \( k = \frac{h^2 - 8h + 20}{4} \) represents a parabola in the \( hk \)-plane.

### Final Result
Thus, the locus of the point which is equidistant from the point (4, 2) and the x-axis is given by the equation:
\[ k = \frac{1}{4}(h^2 - 8h + 20) \]

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