Question 1065506
<pre>
This depends on whether you are still in algebra or in calculus.
I'll do it both ways:

1.  Using calculus:

By implicit differentiation:

{{{4x^2 + y^2 + 8x-2y - 3 = 0}}}

{{{8x+2y*expr((dy)/(dx)) + 8 - 2*expr((dy)/(dx)) = 0}}}

We set {{{(dy)/(dx)}}} equal to 0

{{{8x + 8 = 0}}}

{{{8x=-8}}}

{{{x=-1}}}, substituting:

{{{4(-1)^2 + y^2 + 8(-1)-2y - 3 = 0}}}

{{{4+y^2-8-2y-3=0}}}
{{{y^2-2y-7=0}}}
{{{y = (-(-2) +- sqrt( (-2)^2-4*(1)*(-7)) )/(2*(1)) }}}
{{{y = (2 +- sqrt(4+28) )/2 }}}
{{{y = (2 +- sqrt(32) )/2 }}}
{{{y = (2 +- sqrt(16*2) )/2 }}}
{{{y = (2 +- 4*sqrt(2) )/2 }}}
{{{y = (2(1 +- sqrt(2)))/2 }}}
{{{y = 1 +- sqrt(2) }}}

Answer, since the + gives the larger value, 
the maximum point is {{{(matrix(1,3,-1,",",1+sqrt(2)))}}}

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2.  If you're in algebra,

{{{4x^2 + y^2 + 8x-2y - 3 = 0}}}

Swap the middle two terms to get like lettered terms together:

{{{4x^2 + 8x + y^2 - 2y - 3 = 0}}}

Add 3 to both sides:

{{{4x^2 + 8x + y^2 - 2y = 3 }}}

Factor 4 out of the first two terms:

{{{4(x^2 + 2x) + y^2 - 2y = 3 }}}

Complete the square inside the parentheses:
                {{{2*expr(1/2)=1}}}
                {{{1^2=1}}}
                
Add +1 inside the parentheses and add +4*1 to the right side
since 4 times what is added inside the parentheses is added
to the left side and must also be added to the right side.

{{{4(x^2 + 2x + 1) + y^2 - 2y = 3+4 }}}

Complete the square on the y's:
                {{{-2*expr(1/2)=-1}}}
                {{{(-1)^2=1}}}

Add +1 to both sides:

{{{4(x^2 + 2x + 1) + y^2 - 2y+1 = 3+4+1 }}}

Factor the two trinomials, which are perfect squares, and
write them as the squares of binomials, combine numers
on the right side:

{{{4(x+1)^2 + (y-1)^2 = 8 }}}

Divide through by 8

{{{4(x+1)^2/8 + (y-1)^2/8 = 8/8 }}}

{{{(x+1)^2/2 + (y-1)^2/8 = 1 }}}

Compare to the standard equation for an ellipse:

{{{(x-h)^2/b^2 + (y-h)^2/a^2 = 1 }}} since a > b

center (h,k) = (-1,1), {{{a=sqrt(8)=sqrt(4*2)=2sqrt(2)}}}, {{{b=sqrt(2)}}}

Graph:

{{{drawing(400,400,-2.5,1.5,-1.5,3,graph(400,400,-2.5,1.5,-1.5,3), arc(-1,1,sqrt(2),2sqrt(2)),circle(-1,1,.02),circle(-1,1+sqrt(2),.02),green(line(-1,1,-1,1+sqrt(2))),locate(-1,1,"(-1,1)"),

locate(-1.5,2.7,(matrix(1,3,-1,",",1+sqrt(2))))


 )}}}

The highest point on the ellipse is the upper vertex, which has the
same x-coordinate as the center, which is -1,

The y-coordinate of the upper vertex, which is {{{a=sqrt(2)}}} units 
above the center, which has y-coordinate 1, so we add {{{a=sqrt(2)}}} to
1 and get {{{1+sqrt(2)}}}.

Thus the highest point on the ellipse is:

{{{(matrix(1,3,-1,",",1+sqrt(2)))}}}

Edwin</pre></b>