Question

convert to rectangular form r=4 cos theta

Answer 1

\[r=4\cos\theta\]\[r^2=4r\cos\theta\]\[x^2+y^2=4x\]\[(x-2)^2+y^2=2^2\]

Answer 2

\[
x = r \cos(\theta)
\\
y = r \sin(\theta)\\

\sqrt{x^2+y^2} =r\\

r = 4 \cos(\theta)\\
\sqrt{x^2+y^2} = 4 \frac x r = 4 \frac x {\sqrt{x^2+y^2}}\\
x^2 + y^2 = 4 x \\
\]

Answer 3

eliassaab, what happened to r?

Answer 4

\[ r =\sqrt{x^2+y^2}\]

Answer 5

i mean he r in 4 r/x

Answer 6

I have 4 x/r and not 4 r/x

Answer 7

i meant that, what happened to the r

Answer 8

You cross multiply to get
\[ x^2 + y^2 = 4 x
\]

Answer 9

\[\sqrt{x^2+y^2} = 4 \frac x r = 4 \frac x {\sqrt{x^2+y^2}}\\
\sqrt{x^2+y^2} =4 \frac x {\sqrt{x^2+y^2}}\\
x^2 + y^2 = 4 x
\]

Answer 10

kk, does it have to be in x^2 = y^2 format to be in rectangular form?

Answer 11

It has to be a relation between x and y. In our case, we have the equation of a circle.

Answer 12

how would you solve r^2 = 11 cos 2 theta
would you get X^2 + Y^2 = 11 x/r 2

Answer 13

First you have to use the formula
\[ \cos(2 \theta) = \cos^2(\theta)- sin^2(\theta)
\]

Answer 14

\[
x^2 + y^2 = 11 \left( \frac {x^2}{r^2}- \frac {y^2}{r^2} \right)\\

x^2 + y^2 = \frac {11}{r^2} \left( x^2- y^2 \right)\\
x^2 + y^2 = \frac {11}{x^2+y^2} \left( x^2- y^2 \right)\\
\left( x^2 + y^2 \right)^2 =11 \left( x^2- y^2 \right)\\



\]