A point moves so that its distance from the point $(3, 2)$ is one half of its distance from the line $x = -2$. What is the equation of the curve on which this point is moving?
According to the textbook, the answer is $3x^2 + 4y^2 - 28x - 16y + 48 = 0$. How can I find this equation?
On
Let's consider $M(x,y)$ a point in the plane. It's distance from the line $X=-2$ is $|x+2|$ and it's distance from $F(3,2)$ is $\sqrt{(x-3)^2+(y-2)^2}$. So if $M$ is such that it's distance from $F$ is half the distance to the line we get
$$\sqrt{(x-3)^2+(y-2)^2}={1\over 2}|x+2|$$
Squaring and simplifying one gets the expected result
Consider a point $P(x,y)$ on that line. Then the distance to $x=-2$ equals $\sqrt{(x+2)^2}$, and the distance to $Q(2,3)$ equals $\sqrt{(x-3)^2 + (y-2)^2}$. We know that:
$$\sqrt{(x-3)^2 + (y-2)^2} = \frac{1}{2}\sqrt{(x+2)^2} \iff 4 (x-3)^2 + 4 (y-2)^2 = (x+2)^2$$
$$\iff 3x^2 + 4y^2 - 28x - 16y + 48 = 0$$