Question

convert the polar form r=3acos(alpha) into Cartesian form, and prove it's a circle

Answer 1

\[\begin{align}
r&=3a\cos(\alpha)\\
\therefore \cos(\alpha)&=\frac{r}{3a}\\
\text{and }\sin(\alpha)&=\sqrt{1-\cos^2(\alpha)}\\
&=\sqrt{1-\frac{r^2}{9a^2}}\\
&=\frac{\sqrt{9a^2-r^2}}{3a}\\
x&=r\cos(\alpha)\\
&=\frac{r^2}{3a}\\
\therefore r^2&=3ax\\
y&=r\sin(\alpha)\\
&=\frac{r\sqrt{9a^2-r^2}}{3a}\\
\therefore y^2&=\frac{r^2(9a^2-r^2)}{9a^2}\\
&=\frac{3ax(9a^2-3ax)}{9a^2}
\end{align}\]
can you prove the rest from there?

Answer 2

thank you ! but can you just instruct me how to prove it ?

Answer 3

ok - the final equation can be further simplified as follows...

Answer 4

\[\begin{align}
y^2&=\frac{9a^2x(3a-x)}{9a^2}\\
&=x(3a-x)\\
&=3ax-x^2\\
\therefore x^2+y^2-3ax&=0\\
\end{align}\]now note that \((x-\frac{3a}{2})^2=x^2-3ax+\frac{9a^2}{4}\), which means we have \(x^2-3ax=(x-\frac{3a}{2})^2-\frac{9a^2}{4}\). so, using this we get...

Answer 5

\[\begin{align}
x^2+y^2-3ax&=0\\
\therefore (x-\frac{3a}{2})^2-\frac{9a^2}{4}+y^2&=0\\
\therefore (x-\frac{3a}{2})^2+y^2&=\frac{9a^2}{4}&=(\frac{3a}{2})^2
\end{align}\]

Answer 6

that is the equation of a circle with origin at \((\frac{3a}{2}, 0)\) and radius \frac{3a}{2}

Answer 7

radius = \(\frac{3a}{2}\)

Answer 8

do you understand the steps?

Answer 9

the general equation of a circle with center at (a, b) and radius r is:\[(x-a)^2+(y-b)^2=r^2\]

Answer 10

yes i do, thank you very much .. but one thing, how did you know to do it ? because i tried to do \[x=3acos ^{2}\alpha, y=3acos \alpha \sin \alpha\]
and then do
\[x ^{2}+y ^{2}\]
but then i stayed with the alpha variable

Answer 11

basically you need to find a way of eliminating both r and alpha. this will come with practice as you do more and more examples.

Answer 12

in this case I first got an expression for x interms of just r.
then found y in terms of just r.
and then looked to see how I could eliminate r from the equations.

Answer 13

nice, thank you very much !! i definitely need some more practice with it =]

Answer 14

one approach is to follow this strategy:

1. you are given r interms of alpha.
2. rearrange this to get an expression for cos/sin alpha in terms of r.
3. you know x = rcos(alpha) - so replace cos(alpha) with the outcome of step 2
4. you know y = rsin(alpha) - so replace sin(alpha) with the outcome of step 2
5. rearrange step 3 to get r in terms of x and substitute this into step 4 and voila!

Answer 15

great ! thank you very much !!! saved my homework =]

Answer 16

yw :-)