Q.A point moves such that its distance from the $x$-axis is always equal to its distance from the point $(0,4)$. Find the equation of the locus.

The distance from the $x$-axis is simply $|y|$, and the distance to the fixed point $(0,4)$ uses the ordinary distance formula — equating and squaring the two removes the absolute value and the root together.

Let $P(x,y)$. Condition: $|y| = \sqrt{x^2+(y-4)^2}$. Squaring: $y^2 = x^2+y^2-8y+16$, so $8y = x^2+16$, giving $y=\dfrac{x^2}{8}+2$.

Let $P(x,y)$. Distance of $P$ from the $x$-axis is $|y|$. Distance of $P$ from the point $(0,4)$ is $\sqrt{(x-0)^2+(y-4)^2} = \sqrt{x^2+(y-4)^2}$.

Setting the two distances equal:

Squaring both sides: $y^2 = x^2 + (y-4)^2 = x^2 + y^2 - 8y + 16$.

Cancelling $y^2$ from both sides: $0 = x^2 - 8y + 16$, so:

Independent check. Take $x=0$: the equation gives $y=2$, i.e. the point $(0,2)$. Distance of $(0,2)$ from the $x$-axis is $|2|=2$. Distance of $(0,2)$ from $(0,4)$ is $\sqrt{0+(2-4)^2}=\sqrt4=2$. Both equal $2$ ✓, confirming the derived locus is consistent with the original condition at this point.

Writing the distance from the $x$-axis as $y$ instead of $|y|$ is usually harmless once both sides are squared, but skipping the squaring step and trying to equate $y$ directly with $\sqrt{x^2+(y-4)^2}$ leads nowhere; also watch for a sign slip expanding $(y-4)^2$ as $y^2-4y+16$ instead of the correct $y^2-8y+16$.