Analytic Geometry – Problem 26: Find the locus equation of point

Let the circle $x^2+y^2+2x-15=0$ have center $A$. A line $l$ passes through $B(1,0)$ and is not the $x$-axis. The line $l$ intersects the circle at $C$ and $D$. Through $B$, draw the line parallel to $AC$, meeting $AD$ at point $E$. Find the locus equation of point $E$.

Rewrite the circle: $$(x+1)^2+y^2=16.$$ So its center is $A(-1,0)$ and radius is $r=4$. Hence $$|AC|=|AD|=4.$$

Since $B$ lies on line $l$, points $C,B,D$ are collinear. In triangle $ADC$, the point $B$ is on line $DC$, and the construction gives $BE\parallel AC$ with $E\in AD$. Therefore triangles $\triangle DBE$ and $\triangle DCA$ are similar, so $$\frac{|BE|}{|AC|}=\frac{|DE|}{|DA|}.$$ Because $|AC|=|DA|$, we obtain $|BE|=|DE|$. Then $$|AE|+|BE|=|AE|+|DE|=|AD|=4.$$ So point $E$ satisfies $$|EA|+|EB|=4,$$ which is the definition of an ellipse with foci $A(-1,0)$ and $B(1,0)$, and major axis length $2a=4\Rightarrow a=2$. Here $c=|AB|/2=1$, so $b^2=a^2-c^2=4-1=3$. Thus the locus is $$\frac{x^2}{4}+\frac{y^2}{3}=1.$$ (When $l$ is not the $x$-axis, we exclude the degenerate case on the $x$-axis.)

The locus is the ellipse $$\frac{x^2}{4}+\frac{y^2}{3}=1.$$