1. ## Polar coordinates

The polar equation of a curve C is r^2 = 2 sin 2Ѳ.
(a) Find a Cartesian equation for C.
(b) Show that C could be represented parametrically by the equations

x =(2t)/(1 + t^4), y = (2t^3)/(1+t^4)
where t is a parameter.

I’ve done (a). But how can I solve (b)?

2. Originally Posted by geton
The polar equation of a curve C is r^2 = 2 sin 2Ѳ.
(a) Find a Cartesian equation for C.
(b) Show that C could be represented parametrically by the equations

x =(2t)/(1 + t^4), y = (2t^3)/(1+t^4)
where t is a parameter.

I’ve done (a). But how can I solve (b)?
So you know that $\left( x^2 + y^2 \right)^2 = 4xy$. (Here's some advice: The answer to (a) is needed in order to answer (b). So next time, to save a poor sap like me from spending 10 seconds re-inventing the wheel, be a sport and include your answer).

The easiest approach is to sub the parametric expressions into the cartesian equation and show that you get an identity:

Consider: $x^2 + y^2 = \frac{4t^2}{(1 + t^4)^2} + \frac{4t^6}{(1 + t^4)^2} = \frac{4t^2 (1 + t^4)}{(1 + t^4)^2} = \frac{4t^2}{1 + t^4}$.

Therefore LHS = $(x^2 + y^2)^2 = \frac{16t^4}{\left( 1 + t^4 \right)^2}$.

RHS = $4xy = 4 \left( \frac{2t}{1 + t^4} \right) \left( \frac{2t^3}{1+t^4} \right) = \frac{16t^4}{\left( 1+t^4 \right)^2} = \,$ LHS.

Q.E.D.