Question 746555
An equation of the form
{{{(x-h)^2/(a^2)+(y-k)^2/(b^2)=1}}} represents an ellipse centered at (h,k), with an axis of length {{{2a}}} parallel to the x-axis,
and an axis of length {{{2b}}} parallel to the y-axis.
If we can transform the equation given into such a form, we will be able to find everything the problem asks for.
 
{{{18x^2+4y^2-108x+16y=106}}} --> {{{18x^2-108x+4y^2+16y=106}}} --> {{{(18x^2-108x)+(4y^2+16y)=106}}} --> {{{18(x^2-6x)+4(y^2+4y)=106}}}
At this point, you look at the two expressions in brackets and have to realize that we can add something to each expression to "complete a square"
{{{x^2-6x}}} is part of {{{x^2-6x+9=(x-3)^2}}} and
{{{y^2+4y}}} is part of {{{y^2+4y+4=(y+2)^2}}}
So {{{18(x-3)^2=18(x^2-6x+9)=18x^2-108x+highlight(162)}}} and
{{{4(y+2)^2=4(y^2+4y+4)=4y^2+16y+highlight(16)}}}
So we go back to the original equation, and add {{{162+16}}} to both sides of the equal sign:
{{{18x^2+4y^2-108x+16y=106}}} --> {{{18x^2-108x+162+4y^2+16y+16=106+162+16}}} --> {{{(18x^2-108x+162)+(4y^2+16y+16)=284}}} --> {{{18(x^2-6x+9)+4(y^2+4y+4)=284}}} --> {{{18(x-3)^2+4(y+2)^2=284}}}
Dividing both sides of the equal sign by {{{284}}} the equation turns into
{{{(x-3)^2/((284/18))+(y+2)^2/((284/4))=1}}}
 
That is the equation of an ellipse with {{{highlight(center)}}} at (3,-2).
The axis parallel to the y-axis (along {{{x=3}}}) is longer, and it is called the major axis.
Half of its length (called the semi-major axis) is
{{{sqrt(284/4)=sqrt(71)}}}
The {{{(vertices)}}} are the ends of the major axis, at a distance {{{sqrt(71)}}} from the center, and are at
({{{3}}},{{{-2-sqrt(71)}}}) and ({{{3}}},{{{-2+sqrt(71)}}})
 
The other axis is called the minor axis.
It is along the line {{{y=-2}}}, parallel to the x-axis.
The ends of the minor axis (often called co-vertices) are at distance
{{{sqrt(284/18)=sqrt(142/9)=sqrt(142)/sqrt(9)=sqrt(142)/3}}}
That distance is called the semi-minor axis.
 
An ellipse has two {{{highlight(foci)}}} located on the major axis, between the center and the vertices, at a distance from the center {{{c}}} called the focal distance. That distance {{{c}}}, and the semi-minor axis are the legs of a right triangle with the semi-major axis for a hypotenuse.
Applying the Pythagorean theorem, we find that
{{{c^2+142/9=71}}} --> {{{c^2=71-142/9}}} --> {{{c^2=71-142/9}}} --> {{{c^2=71-142/9}}} --> {{{c^2=497/9}}} --> {{{c=sqrt(497/9)}}} --> {{{highlight(c=sqrt(497)/3)}}}
So the {{{highlight(foci)}}} are at
({{{3}}},{{{-2-sqrt(497)/3}}}) and ({{{3}}},{{{-2+sqrt(497)/3}}})
{{{drawing(200,400,-2,8,-12,8,
grid(0), circle(3,-2,0.2),
line(3,-2-sqrt(71),3,-2+sqrt(71)),line(-0.972,-2,6.972,-2),
red(circle(3,-2-sqrt(71),0.2)),red(circle(3,-2+sqrt(71),0.2)),
green(circle(-0.972,-2,0.2)), green(circle(6.972,-2,0.2)), 
blue(circle(3,-9.431,0.2)), blue(circle(3,5.431,0.2)),
locate(3.3,-10,vertex),locate(3.3,7,vertex),
locate(3.3,-9,focus),locate(3.3,6,focus),locate(3.2,-2.1,center)
)}}}  {{{graph(200,400,-2,8,-12,8,-sqrt(284-18(x-3)^2)/2-2,sqrt(284-18(x-3)^2)/2-2)}}}