Finding the locus of the following conditions Find the locus of all points in the plane, whose distance from a constant point $F=(x_0,y_0)$ divided by their distance from the vertical line $L=\{(k,y)\mid y \in \mathbb R \}$ equals a constant $q<1$. Meaning, we need finding all points that satisfy $E=\{p\in \mathbb R^2\mid {d(p,F) \over \ d(p,L)}=q \}$
Thanks in advance for any assistance!
 A: We assume that $F=(0,0)$. (Otherwise, you can shift the plane over a distance $-(x_0,y_0)$.) $L$ is the vertical line with $x=k$. For a point $P=(x,y)$, we know that $d(P,F)=\sqrt{x^2+y^2}$. Also, $d(P,L)=\sqrt{(x-k)^2}=|x-k|$. Now, we want
$$
\frac{\sqrt{x^2+y^2}}{\sqrt{(x-k)^2}}=q
$$
We are going to solve this for $y$:
$$
x^2+y^2=q^2(x-k)^2
$$
Thus,
$$
y=\pm \sqrt{q^2(x-k)^2-x^2}
$$
Now, the set $E$ is the following:
$$
E=\left\{(x,y)\in\mathbb R^2\left|\,|y|=\sqrt{q^2(x-k)^2-x^2}\right.\right\}
$$
We can also write this as
$$
(1-q^2)x^2-2q^2kx+y^2=q^2k^2
$$
(this follows from the second equation above.)
This can be rewritten to
$$
(1-q^2)\left(x-\frac{q^2}{1-q^2}k\right)^2+y^2=q^2k^2+\left(\frac{q^2}{1-q^2}k\right)^2
$$
This is the general formula for a conic section. Because $q<1$, the factors in front of the $x$ and $y$ are both positive. The right hand side is positive too. Therefore, this is the general formula of an ellipse with center
$$(x,y)=\left(\frac{q^2}{1-q^2}k,0\right)$$
