Mathematical Immaturity

13.25. Miscellaneous review exercises on conic sections

• Consider the locus of the points $P$ in the plane for which the distance of $P$ from the point $(2, 3)$ is equal to the sum of the distances of $P$ from the two coordinate axes. (a) Show that the part of this locus which lies in the first quadrant is part of a hyperbola. Locate the asymptotes and make a sketch.
  (b) Sketch the graph of the locus in the other quadrants.
• If we set the focus $F = (2, 3),$ then the distance $\|P - F \| = x + y.$ This means for some eccentricity $e > 0,$ $$x + y = ed(P, L)$$where $d(P, L)$ is the distance from $P$ to the directrix line $L.$
• (a) Let the focus $F = (2, 3),$ then the distance $\|P - F \| = x + y.$ Then, for some directrix $L$ and eccentricity $e > 0,$ $$x + y = ed(P, L)$$where $d(P, L)$ is the distance from $P$ to $L.$ But by definition, if $N$ is the vector normal to $L$ and $X$ is a point on $L,$ $d(P, L) = \frac{|(P - X) \cdot N|}{\|N\|}.$ Applying this to our previous equation we see that $$\frac{x + y}{e} = \frac{|(P - X) \cdot N|}{\|N\|}$$But since $P = (x, y)$ if we denote $X = (x_0, y_0),$ this means that $(P - X) \cdot N = (x - x_0, y - y_0)\cdot N = (x + y),$ which is satisfied by $X = O$ and $N = (1, 1).$ But if $N = (1, 1),$ then $\|N\| = \sqrt{2},$ thus $e$ must also be $\sqrt{2}$ if the above equation is to be satisfied. Thus, the locus of points $P$ in $Q1$ whose distance from $F = (2,3)$ is equal to $(x + y)$ must be part of a hyperbola since $e > 1.$ To find the asymptote lines of the hyperbola, recall that $\|P - F\| = x + y,$ which means that if we square both sides we get $$2xy + 4x + 6y = 13$$Put in terms of $y$ we get $$y = \frac{13 - 4x}{2x + 6}$$As $x$ approaches either $+\infty$ or $-\infty,$ we see that the limit of $y$ goes to $-2.$ As $x$ approaches $-3$ from the left, we see that $y$ goes to $-\infty$ and likewise $y$ goes to $+\infty$ as $x$ approaches $-3$ from the right. Thus, the asymptote lines of the hyperbola are the lines $x = -3$ and $y = -2$