Question 170283
Remember, the standard equation for any ellipse is {{{((x-h)^2)/(a^2)+((y-k)^2)/(b^2)=1}}}



Also, the foci of any ellipse will lie on the major axis. 


Since EVERY point given to you has a y-coordinate of 0, this means that the major axis lies horizontally. So this tells us that the length of the major axis is {{{2a}}} (since "a" corresponds with the horizontal axis)



Finding the center:



To find the center, simply average the vertices to get



x-mid: {{{(-6+6)/2=0/2=0}}}


So the x-coordinate of the center is 0 which means that h=0




y-mid: {{{(0+0)/2=0/2=0}}}


So the 0-coordinate of the center is 0 which means that k=0



So the center is (0,0) which tells us that h=0 and k=0



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Finding the length of the Major Axis:



Now let's find the distance from the two vertices:



Distance: {{{d=sqrt((-6-6)^2+(0-0)^2)=sqrt((-12)^2+0^2)=sqrt(144)=12}}}



So the distance between the two vertices is 12 units (note: you can just count the units in between by use of a graph). This means that the length of the major axis is 12 units.



So this means that {{{2a=12}}} which tells us that {{{a=6}}}




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Finding the length of the Minor Axis:



To find the length of the minor axis (ie find the value of "b"), we first need to find the distance from the center to either focus


That distance is


Distance: {{{d=sqrt((0-2)^2+(0-0)^2)=sqrt((-2)^2+(0)^2)=sqrt(4)=2}}}



So the distance from the center to either focus is 2 units. This means that {{{c=2}}}



Now the value of "b" can be found through the formula



{{{a^2=b^2+c^2}}}



{{{6^2=b^2+2^2}}} Plug in {{{a=6}}} and {{{c=2}}}



{{{36=b^2+4}}} Square each value.



{{{36-4=b^2}}} Subtract 4 from both sides.



{{{b^2=32}}} Subtract and rearrange the equation.



{{{b=sqrt(32)}}} Take the square root of both sides. Note: only the positive square root is considered



{{{b=4*sqrt(2)}}} Simplify the square root.




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So the values we've found was {{{h=0}}}, {{{k=0}}}, {{{a=6}}}, and {{{b=4*sqrt(2)}}}



{{{((x-h)^2)/(a^2)+((y-k)^2)/(b^2)=1}}} Go back to the general equation of an ellipse



{{{((x-0)^2)/(6^2)+((y-0)^2)/((4*sqrt(2))^2)=1}}} Plug in {{{h=0}}}, {{{k=0}}}, {{{a=6}}}, and {{{b=4*sqrt(2)}}}




{{{((x-0)^2)/(36)+((y-0)^2)/(16*2)=1}}} Square 6 to get 36. Square {{{4*sqrt(2)}}} to get {{{16*2}}}



{{{((x-0)^2)/(36)+((y-0)^2)/(32)=1}}} Multiply



{{{(x^2)/(36)+(y^2)/(32)=1}}} Simplify



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Answer:




So the equation of the ellipse with foci of (-2,0) and (2,0) and vertices (-6,0) and(6,0) is {{{(x^2)/(36)+(y^2)/(32)=1}}}