Q.Find the equation of the locus of a point which is equidistant from the point $S(3,0)$ and the line $x = -3$.

This is the focus-directrix condition that defines a parabola: the point's distance to $S(3,0)$ must equal its perpendicular distance to the vertical line $x=-3$.

Let $P(x,y)$. Distance to $S(3,0)$ is $\sqrt{(x-3)^2+y^2}$; perpendicular distance to $x=-3$ is $|x+3|$. Setting them equal and squaring: $(x-3)^2+y^2=(x+3)^2$, which simplifies to $y^2=12x$.

Let $P(x,y)$ be the moving point. Its distance from $S(3,0)$ is $\sqrt{(x-3)^2+y^2}$. Its perpendicular distance from the vertical line $x=-3$ is $|x-(-3)| = |x+3|$.

Setting these equal (the given condition):

Squaring both sides (valid since both sides are non-negative):

Expanding: $x^2-6x+9+y^2 = x^2+6x+9$.

Cancelling $x^2$ and $9$ from both sides: $-6x+y^2 = 6x$, so:

Independent check. Pick a point that should lie on this locus, say where $x=3$: then $y^2=36$, so $y=\pm6$; take $(3,6)$. Distance from $(3,6)$ to $S(3,0)$ is $\sqrt{0+36}=6$. Perpendicular distance from $(3,6)$ to the line $x=-3$ is $|3+3|=6$. Both distances equal $6$ ✓, confirming the derived equation is consistent with the original condition, independently of the algebra used to derive it.

Forgetting the absolute value when writing the perpendicular distance to a vertical line $x=k$ as $|x-k|$ (not simply $x-k$) can introduce a sign error, though squaring both sides usually absorbs it — the more serious mistake is expanding $(x+3)^2$ incorrectly as $x^2+9$ (dropping the middle term $6x$), which would wrongly cancel everything and give $y^2=0$.