Question 339645
I don't understand how my book comes up witht these answers: If anyone can help, I'd greatly appreciate it!

Convert from polar equation to rectangular equation:
Number 1
<pre><font size = 3 color = "indigo"><b>
{{{r=4/(1+2Sin(theta))}}}

Draw this picture which defines x,y,r, and <font face = "symbol">q</font>
and all their relationships.

{{{drawing(300,180,-.5,.5,-.3,.3, graph(300,180,-.5,.5,-.3,.3),
triangle(0,0,.35,0,.35,.2), locate(.11,.06,theta), locate(.19,0,x),
locate(.36,.12,y), locate(.15,.14,r)
 )}}}

{{{r=4/(1+2Sin(theta))}}}

Clear of fractions:

{{{r(1+2Sin(theta))=4}}}

Replace any trig functions:

Refer to the picture above and you can see that you can 
replace {{{Sin(theta)}}} by {{{y/r}}}

{{{r(1+2*expr(y/r))=4}}}

{{{r+2y=4}}}

Now replace the r by using the Pythagorean theorem {{{r=sqrt(x^2+y^2)}}}
from the picture above.

{{{sqrt(x^2+y^2)+2y=4}}}

Isolate the radical term:

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

Square both sides:

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

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

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

{{{x^2+y^2=16-8y-8y+4y^2}}}

{{{x^2+y^2=16-16y+4y^2}}}

{{{x^2-3y^2 + 16y -16 = 0}}}

-------------------------------

Also, this one...they ask to change it to a polar equation.

{{{x^2 - y^2 = 12}}}

Refer to the picture above, write the expressions for the sine and
the cosine:

{{{x/r=cos(theta)}}},  {{{y/r=sin(theta)}}}

Solve for x and y:

{{{x=r*cos(theta)}}},  {{{y=r*sin(theta)}}}

Substitute these:

{{{(r*cos(theta))^2 - (r*sin(theta))^2 = 12}}}

{{{r^2*Cos^2theta - r^2*Sin^2theta = 12}}}

Factor out {{{r^2}}} on the left

{{{r^2(Cos^2theta-Sin^2theta)=12}}}

Now we can recognize that what's in the parentheses is the right side 
of the identity {{{Cos(2theta)=Cos^2theta-Sin^2theta}}} which you
should have memorized. So substituting:

{{{r^2(Cos(2theta))= 12

Solve for {{{r^2}}}

{{{r^2=12/(Cos(2theta))}}}

Replace the {{{Cos(2theta)}}} by {{{1/Sec(2theta)}}}

{{{r^2=12/(1/Sec(2theta)))}}}

Invert and multiply:

{{{r^2=12*Sec(2theta)))}}}

Edwin</pre>