A point moves so that it is equidistant from the points $(1,4)$ and $(5,2)$. Find the equation of its locus.
Since the point $P(x,y)$ is equidistant from $(1,4)$ and $(5,2)$:
$$ \sqrt{(x-1)^2 + (y-4)^2} = \sqrt{(x-5)^2 + (y-2)^2} $$ Square both sides: $$ (x-1)^2 + (y-4)^2 = (x-5)^2 + (y-2)^2 $$ Expand: $$ x^2 - 2x + 1 + y^2 - 8y + 16 = x^2 - 10x + 25 + y^2 - 4y + 4 $$ Combine like terms: $$ 10x - 2x - 8y + 4y - 12 = 0 $$ Simplify: $$ 8x - 4y - 12 = 0 $$ Divide by 4: $$ 2x - y - 3 = 0 $$The locus is the line:
$$ \boxed{2x - y - 3 = 0} $$Point $P(x,y)$ moves so that its distance from the point $(0,1)$ is one half of its distance from the line $y=4$. Compute the eccentricity of the resulting curve.
The condition is:
$$ d_1 = \tfrac12 d_2 $$ Distance to the point $(0,1)$: $$ d_1 = \sqrt{(x-0)^2 + (y-1)^2} $$ Distance to the line $y=4$: $$ d_2 = |4 - y| $$ Impose the condition: $$ \sqrt{(x)^2 + (y-1)^2} = \tfrac12 (4 - y) $$ Square both sides: $$ x^2 + (y-1)^2 = \tfrac14 (4 - y)^2 $$ Expand: $$ x^2 + y^2 - 2y + 1 = \tfrac14 (16 - 8y + y^2) $$ Multiply both sides by 4: $$ 4x^2 + 4y^2 - 8y + 4 = 16 - 8y + y^2 $$ Simplifying, we obtain the equation of an ellipse. $$ 4x^2 + 3y^2 = 12 $$ Divide by 12: $$ \frac{x^2}{3} + \frac{y^2}{4} = 1 $$ Thus the ellipse has center $(0,0)$, with: $$ a = 2,\quad b = \sqrt{3} $$ Compute $c$: $$ c^2 = a^2 - b^2 = 4 - 3 = 1 $$ $$ c = 1 $$ Eccentricity: $$ e = \frac{c}{a} = \frac12 $$Therefore, the eccentricity is:
$$ \boxed{\frac12} $$