Question 1157498


given:
 the equation of the ellipse {{{(x+1)^2/4+(y-1)^2/1=1 }}}
reflected in the line {{{y=-x+12}}}

from {{{(x+1)^2/4+(y-1)^2/1=1}}}, we see that center of the ellipse is at

({{{-1}}},{{{1}}})

since the center of the new ellipse reflected in the line will lie on the line which is perpendicular to given line, find the equation of the line perpendicular to given line and passes through ({{{-1}}},{{{1}}})


{{{y=mx+b}}}

perpendicular line have a slope negative reciprocal to the slope of given line, 
so {{{m=-1/-1=1}}}

{{{y=1x+b}}}
use point ({{{-1}}},{{{1}}}) to find {{{b}}}

{{{1=1(-1)+b}}}

{{{1=-1+b}}}

{{{1+1=b}}}

{{{b=2}}}

perpendicular line is {{{y=x+2}}}


now find intersection of these two lines

{{{-x+12=x+2}}}

{{{-2+12=x+x}}}

{{{2x=10}}}

{{{x=5}}}

{{{y=5+2=7}}}

intersection is at

({{{5}}},{{{7}}})


 the point  ({{{5}}},{{{7}}}) is {{{mid-point}}} between the center to ({{{-1}}},{{{1}}}) and  the new center of the reflected ellipse ({{{x}}},{{{y}}})  on the other side of the given line 

so, ({{{5}}},{{{7}}})=({{{x/2}}},{{{y/2}}})=> ({{{x}}},{{{y}}})=({{{10}}},{{{14}}})

and, equation of the reflected ellipse is

{{{(x-10)^2/4+(y-14)^2/1=1 }}}

graph:

{{{ graph( 600, 600, -20, 20, -20, 20,-x+12,-sqrt(1-(x+1)^2/4)+1,sqrt(1-(x+1)^2/4)+1 ,sqrt(1-(x-10)^2/4)+14, -sqrt(1-(x-10)^2/4)+14) }}}