Is this right?

[kaebun]

polar -> cartesien I think I've got this figured out but the answer I got is weird and since I was so dumb last time I just wanted to make sure I did this right...

$\displaystyle r=15/(2+5cost)$
$\displaystyle 2r+5rcost=15$
$\displaystyle 2sqrt(x^2+y^2)+5x=15$
$\displaystyle 2sqrt(x^2+y^2)=15-5x$
$\displaystyle (2sqrt(x^2+y^2)=15-5x)^2$
$\displaystyle 4(x^2+y^2)=(15-5x)^2$
$\displaystyle 4x^2+4y^2=25x^2-150x+225$
$\displaystyle 4x^2+4y^2-25x^2+150x=225$
-21x^2+150x+4y^2=225
-21(x^2+150/-21x+ 204.1)+4y^2=225+204.1
-21(x^2+300/-21)^2)+4y^2=429.1
-21(x+300/-21)^2+4y^2=429.1
$\displaystyle y^2/107.28-(x-300/21)^2/20.4=1$

 

[tkhunny]

$\displaystyle (2sqrt(x^2+y^2)=15-5x)^2$

Excellent example of confusing and meaningless notation.

Why did you switch to decimals? That just isn't helpful. It looks fine until you did that, then it becomes confusing.

 

[Gene]

I agree up to
4x^2+4y^2=25x^2-150x+225
I would have gone with
4y^2=21(x^2-150/21x)+225
4y^2=21(x^2-50/7x+(25/7)²)+225-21(25/7)²
4y^2=21(x-(25/7))²+225-21*(25/7)²
You have an error in your square completion. You should add (b/2)² not 2b²
Going decimal I get
4y^2=21(x-3.57)²-42.86
No less weird but...
Finish it up. Hint When x=5 y=0