Question 111547
what are 2 functions for the following equation: 
{{{((x+1)/10)^2 + (y/5)^2 = 1}}}
<pre><font size = 4><b>
This is the equation of this ellipse:

{{{drawing(500,234.375,-12,10,-5.5,5.5,
   graph(500,234.375,-12,10,-5.5,5.5,5sqrt(1-((x+1)/10)^2)), 
   graph(500,234.375,-12,10,-5.5,5.5,-5sqrt(1-((x+1)/10)^2)) 
 )}}}

An ellipse is NOT a function because it does not pass the
vertical line test.

However the upper half of the ellipse is a function:

{{{drawing(500,234.375,-12,10,-5.5,5.5,
   graph(500,234.375,-12,10,-5.5,5.5,5sqrt(1-((x+1)/10)^2)) 
 )}}}

Also the lower half of the ellipse is a function:

{{{drawing(500,234.375,-12,10,-5.5,5.5,
   graph(500,234.375,-12,10,-5.5,5.5,-5sqrt(1-((x+1)/10)^2)) 
 )}}}

To find these two functions we solve the equation for y:

{{{((x+1)/10)^2 + (y/5)^2 = 1}}}
{{{(x+1)^2/10^2 + y^2/5^2 = 1}}}
{{{(x+1)^2/100 + y^2/25 = 1}}}
Multiply through by LCD = 100
{{{(x+1)^2 + 4y^2 = 100}}}
{{{4y^2 = 100-(x+1)^2}}}
{{{4y^2 = 100-(x^2+2x+1)}}}
{{{4y^2 = 100-x^2-2x-1}}}
{{{4y^2 = 99-x^2-2x}}}
{{{y^2 = (99-x^2-2x)/4}}}
{{{y}}} = ±{{{sqrt(99-x^2-2x)/2}}}

When we use the positive sign we get the
functional equation

{{{y}}} = {{{sqrt(99-x^2-2x)/2}}}

which is the equation of the top half of
the ellipse.

When we use the negative sign we get the
functional equation

{{{y}}} = {{{-sqrt(99-x^2-2x)/2}}}

which is the equation of the bottom half of
the ellipse.

Edwin</pre>