Equation relating square of distance to point and distance to line

Question

Find the equation of the locus of the point $P(x, y)$ such that the square of the distance from $(-2, -5)$ to $P(x, y)$ is three times the distance from $P(x, y)$ to the line $8x+15y=34$.

My answer:

$ y=-(8/15)x + (34/15)$

$ ((x+2)^2+(y+5)^2 = 3((y+(8/15)x + (34/15)) $

$ x^2+4x+4+y^2+10y+25 = 3y + (8/5)x - (34/5) $

$ 5x^2+5y^2-4x-5y+179 = 0 $

The answer given was: $17X^2+17y^2+92x+215y+391 = 0$.

What accounts for the difference in my answer and the given one?

Answer 1

Let's start with the normal form of the equation of the line:

$$ 8x + 15y = 34 $$

Since $8^2 + 15^2 = 17^2$, the normal form is:

$$ \frac{8}{17} x + \frac{15}{17} y = 2 $$

and the distance from point $P(x,y)$ to this line is:

$$ \left | \frac{8}{17} x + \frac{15}{17} y - 2 \right | $$

Note, for example, that a point on the line will give a distance zero in the last expression.

Thus we can express the locus of points whose distance squared to $(-2,-5)$ is three times the distance to the given line by:

$$ (x+2)^2 + (y+5)^2 = 3 \left | \frac{8}{17} x + \frac{15}{17} y - 2 \right | \tag{1}$$

Having an equation to work with is only the starting point for analyzing the curve or curves made of points satisfying this. We might well ask, are there such points on both sides of the line, or only on one side? To the extent that the Question asks about reconciling two (possible) equations, neither of which is precisely like the above, we need to pursue the analysis a little bit.

To begin, the point $(-2,-5)$ is not on the line. In fact, its distance from the point to the line is:

$$ \left | \frac{8}{17} (-2) + \frac{15}{17} (-5) - 2 \right | = \frac{125}{17} $$

From this fact we can deduce that the only points satisfying the equation are on the same side of the line as $(-2,-5)$. A sketch of an argument is this:

Suppose a point lies at distance $d \gt 0$ from the line but on the side opposite from $(-2,-5)$. Then its distance to $(-2,-5)$ is at least $d + \frac{125}{17}$. That distance squared is then at least:

$$ \left( d + \frac{125}{17} \right)^2 = d^2 + \frac{250}{17} d + \left( \frac{125}{17} \right)^2 $$

Since $\frac{250}{17} \gt 3$, it is evident that the distance to $(-2,-5)$ squared will exceed $3d$, and thus no points on the "far side" of the line will belong to the locus.

This allows us to eliminate the absolute value signs in (1), knowing that for points in the locus the expression inside the absolute value is always negative, as it would be for $(-2,-5)$ and thus for all points on its same side of the line. Therefore we can write instead:

$$ (x+2)^2 + (y+5)^2 = -3 \left( \frac{8}{17} x + \frac{15}{17} y - 2 \right) \tag{2}$$

Multiply both sides by $17$ to clear fractions:

$$ 17x^2 + 68x + 17y^2 + 170y + 17(4+25) = -24x - 45y + 17\cdot 6 \tag{3} $$

With slightly more simplification we have just the answer "given" to you:

$$ 17x^2 + 17y^2 + 92x + 215y + 391 = 0 \tag{4} $$