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Consider the parabola given by the equation y^2-2x-6y+11=0
How to get it in Standard Form and find the vertex and focus

Find the center of the ellipse
5x^2+2y^2-20x+24y+82=0

Find the length of the curve over the given interval
r=8(1+cos(angle), 0<angle<2(3.124)

Find the area of the region.
interior of r=5(1+sin(angle))

and

One petal of r=cos(5(angle))

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Originally Posted by Music4Life

Find the area of the region.
interior of $r=5(1+sin(t))$

and

One petal of $r=cos(5t)$

Is this one problem or two? I am going to treat it as two different regions.

$\frac{25}{2}\int_{0}^{2\pi}(1+sin(t))^{2}dt$

$=\frac{25}{2}\int_{0}^{2\pi}\left[2sin(t)dt-\frac{1}{2}cos(2t)+\frac{3}{2}\right]dt$

Now, integrate away.

As for the rose, to find the limits of integration, solve $cos(5t)=0$

We find $t=\frac{\pi}{10}$

Multiply by 2 because of symmetry.

$\int_{0}^{\frac{\pi}{10}}(cos(5t))^{2}dt=\int_{0}^ {\frac{\pi}{10}}\left[\frac{1+cos(10t)}{2}\right]$

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Find the length of the curve over the given interval
$r=8(1+cos(t)), \;\ 0<t<2{\pi}$

$r=8(1+cos(t))$

Polar arc length is given by:

$\int_{\alpha}^{\beta}\sqrt{r^{2}+\left(\frac{dr}{d t}\right)^{2}}dt$

$\frac{dr}{dt}=-8sin(t)$

I assume you are integrating from 0 to $2{\pi}$

$\int_{0}^{2\pi}\sqrt{(8(1+cos(t)))^{2}+(-8sin(t))^{2}}dt$

$\int_{0}^{2\pi}\sqrt{128(cos(t)+1)}dt$

This integral can be a little tricky. As a matter of fact, most calculators grunt trying to do it, but it ain't that bad if we use some tricks.

$8\sqrt{2}\int_{0}^{2\pi}\sqrt{1+cos(t)}dt$

$\sqrt{\frac{1+cos(t)}{1}\cdot\frac{1-cos(t)}{1-cos(t)}}$

$=\sqrt{\frac{1-cos^{2}(t)}{1-cos(t)}}$

$=\sqrt{\frac{sin^{2}(t)}{1-cos(t)}}$

$=\frac{sin(t)}{\sqrt{1-cos(t)}}$

Now, let $u=1-cos(t), \;\ du=sin(t)dt$

We end up with something easy:

$\int\frac{1}{\sqrt{u}}du=2\sqrt{u}$

Resub:

$2\sqrt{1-cos(t)}$

Let's integrate from 0 to Pi and multiply by 2 because using 2Pi presents a conundrum. Do you see why?

$2\cdot 8\sqrt{2}\sqrt{1-cos(\pi)}=64$

There it is.

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Find the center of the ellipse
$5x^{2}+2y^{2}-20x+24y+82=0$

Complete the square.

$5x^{2}-20x+2y^{2}+24y=-82$

$5(x^{2}-4x+4)+2(y^{2}+12y+36)=-82+20+72$

$5(x-2)^{2}+2(y+6)^{2}=10$

Now, can you finish and get into the form of an ellipse? Just one more step.

Though, you can easily see the center coordinates as it is now.

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Consider the parabola given by the equation $y^{2}-2x-6y+11=0$
How to get it in Standard Form and find the vertex and focus

As with the ellipse, complete the square.

The standard form of this parabola is $x=a(y-h)^{2}+k$ , where (h,k) are the vertex coordinates.

It is a parabola opening to the right symmetric about y=3. The directrix is at x=1/2.

$y^{2}-6y-2x=-11$

$(y^{2}-6y+9)-2x=-11+9$

$(y-3)^{2}+2=2x$

$x=\frac{1}{2}(y-3)^{2}+1$

The focus can be found by using $p=\frac{1}{4a}$ .