Consider the locus of the points X in the plane for wich the distance of X from the point (2,3) is equal to the sum of the distances of X from the coordinate axes. a) Show that the part of this locus which lies in the first quadrant is part of a hyperbola.locate the assymptotes and make a sketch. b) Sketch the graph of the locus in the other quadrants.
ATTEMPT:a) Let F=(2,3),Then for the first quadrant $\Vert X-F \Vert =x+y= \sqrt 2 (\frac {x-2}{\sqrt 2}+\frac {y-3}{\sqrt 2} +\frac{5}{\sqrt 2})$ This is a vector equation of a hyperbola of the form $\Vert X-F \Vert =e\vert (X-F)\cdot N+d\vert$.So $e=\sqrt 2$,$X=(x,y)$ and $d=\frac{5}{\sqrt 2}$(d is the distance from the focus F to the directrix),$N=(\frac {1}{\sqrt 2},\frac {1}{\sqrt 2})$ is the unit normal to the directrix L.F must lie on the possitive part of the hyperbola determined by N.The transverse axis of the hyperbola has the same direction with N and includes F,So the equation of the axis must be x=y+1.If we choose an X wich lies on the axis and using the relations $\vert X-F \Vert=e d(X,L)$ and $ d=d(X,L)+\Vert X-F\Vert $ then we can find this X from $d=\frac {\Vert X-F \Vert}{e}+\Vert X-F\Vert$.Now to find the center C of the hyperbola we use $(\Vert F-C \Vert -d)e=\Vert X-C \Vert $ (X is the X wich lies on the axis). So $\vert \alpha \vert =\Vert X-C \Vert$ and $b=\vert \alpha \vert \sqrt {e^2-1}$.The assymptotes pass through the C and are the hypotenuses of the two triangles with sides $\vert \alpha \vert$ and b on the two planes determined by the transverse axis.
QUESTIONS: A)Is the solution correct; As far as the b) part of the exercise I found $N=(\frac {-1}{\sqrt 2},\frac {1}{\sqrt 2})$ and $d=\frac {1}{\sqrt 5}$It seems the hyperbola described by this equation has most of its left part in the first quadrant.Propably i must sketch only the points lying in the second quadrant.?If i got the hyperbola correctly.
I think you are overcomplicating the question. Notice first of all that point $X$ exists only if $x+y\ge0$, i.e. if $X$ lies above the line of equation $y=-x$. In this case we can square equation $||X-F||=x+y$ to get $$ (x-2)^2+(y-3)^2=(x+y)^2, $$ which is the same as $$ (x+3)(y+2)={25\over2}. $$ The latter is the equation of an equilateral hyperbola, with center at $(-3,-2)$ and asymptotes parallel to the coordinate axes. Its upper branch all lies in the region above line $y=-x$ and is the required locus, while its lower branch lies below that line and must be discarded.