Question 838920
Sum of two distances must be 8 units.


Distance from P to (-2,0) PLUS Distance from P to (4,0) EQUALS 8.


You use the Distance Formula.


{{{sqrt((x-(-2))^2+(y-0)^2)+sqrt((x-4)^2+(y-0)^2)=8}}}
{{{sqrt((x+2)^2+y^2)+sqrt((x-4)^2+y^2)=8}}}
{{{sqrt((x+2)^2+y^2)=8-sqrt((x-4)^2+y^2)}}}
Square both sides;
{{{(x+2)^2+y^2=64-16*sqrt((x-4)^2+y^2)+(x-4)^2+y^2}}}
{{{(x+2)^2=-16*sqrt((x-4)^2+y^2)+(x-4)^2+64}}}
{{{(x+2)^2-(x-4)^2-64=-16*sqrt((x-4)^2+y^2)}}}
{{{x^2+4x+4-(x^2-8x+16)-64=-16*sqrt((x-4)^2+y^2)}}}
---steps done on papter---
{{{256y^2+112x^2-1136x=1680}}}, and need to complete the square process for terms with x, not showing the process here...
{{{256y^2+112(x-71/14)^2=1680}}}
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{{{highlight((1/(105/7))(x-71/14)^2+(1/(105/16))(y)^2=1)}}}
This is the standard form of an ellipse.  
{{{a^2=105/7}}}  and {{{b^2=105/16}}}.
The center is not translated along the y axis, but is translated along the x axis, {{{71/14}}} units to the right.