- Translate each inequality into rectangular coordinates. Assume
*r* is non-negative.%Graph the region being described by the inequality.

- sin θ < 1/r
- rcos
^{2}θ + rsin^{2}θ = 1 - 5r
^{2}cos^{2}θ + 10rcosθ + 5 -4r^{2}sin^{2}θ + 16rsinθ-16 = 20

Hint

substitute in *x* and *y* before trying to rearrange

Answer

\item Multiply both sides by *r* to find

sin θ &<& \frac{1}{r}rsinθ&<&1y&<&1

\item Factor out *r* on the left-hand side:

rcos^{2}θ + rsin^{2}θ& = &1r(cos^{2}θ + sin^{2}θ)& = &1r& = &1

Then square both sides:

r& = &1r^{2}& = &1x^{2} + y^{2}& = &1

\item Substitute in *x* and *y* first:

5r^{2}cos^{2}θ + 10rcosθ + 5 -4r^{2}sin^{2}θ + 16rsinθ-16& = &205({ rcosθ})^{2} + 10{ rcosθ} + 5-4({ rsinθ})^{2} + 16{ rsinθ}-16& = &205{ x}^{2} + 10{ x} + 5-4{ y}^{2} + 16{ y}-16& = &20.

Now we can factor to tidy things.

5x^{2} + 10x + 5-4y^{2} + 16y-16& = &205(x^{2} + 2x + 1)-4(y^{2}-4y + 4)& = &205(x + 1)^{2}-4(y-2)^{2}& = &20.

Dividing both sides by 20, we see that this is the equation of a hyperbola:\frac{(x + 1)^{2}}{4}-\frac{(y-2)^{2}}{5} = 1.