Step 1: Analyzing the Locus Definition:
Wait, the problem statement says "distance of 5 units from a fixed point... and also from a fixed line". This phrasing is slightly ambiguous. Usually, "distance from focus = distance from directrix" defines a parabola. Here, if the distance from the point equals the distance from the line (both being variable but equal to each other), it's a parabola. If it means the point is at distance 5 from the point AND distance 5 from the line, that's just the intersection of a circle and parallel lines (points).
However, looking at the options (quadratic equations in x, y), this represents a conic section, likely a parabola where the eccentricity \( e=1 \). The "distance of 5 units" might be a typo for "equidistant" or the distance itself is a parameter irrelevant to the shape type if we assume the standard definition: Distance from Focus \( (1,4) \) = Distance from Directrix \( 2x+3y-1=0 \).
Step 2: Equation of Parabola:
Let \( P(x,y) \) be the point.
Distance from Focus \( S(1,4) \): \( \sqrt{(x-1)^2 + (y-4)^2} \)
Distance from Line \( 2x+3y-1=0 \): \( \frac{|2x+3y-1|}{\sqrt{2^2+3^2}} = \frac{|2x+3y-1|}{\sqrt{13}} \)
Equating squares:
\[ (x-1)^2 + (y-4)^2 = \frac{(2x+3y-1)^2}{13} \]
\[ 13(x^2 - 2x + 1 + y^2 - 8y + 16) = (2x+3y-1)^2 \]
Expand RHS \( (2x+3y-1)^2 = 4x^2 + 9y^2 + 1 + 12xy - 4x - 6y \).
Multiply LHS by 13:
\[ 13x^2 + 13y^2 - 26x - 104y + 221 \]
Equate and rearrange:
\[ 13x^2 + 13y^2 - 26x - 104y + 221 = 4x^2 + 9y^2 + 12xy - 4x - 6y + 1 \]
\[ (13-4)x^2 - 12xy + (13-9)y^2 + (-26+4)x + (-104+6)y + (221-1) = 0 \]
\[ 9x^2 - 12xy + 4y^2 - 22x - 98y + 220 = 0 \]