Transformations of conics?

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May 15th 2011, 07:55 PM
homeylova223

Transformations of conics?

Write an equation of the translated or rotated graph in general form.

y=3x^2-2x+5 T(2,-3)

I know the graph is a parabola. But I am not sure where I would "plug in" h and k into this equation. To get the translated equation.
May 16th 2011, 12:47 AM
earboth

Quote:

Originally Posted by homeylova223

Write an equation of the translated or rotated graph in general form.

y=3x^2-2x+5 T(2,-3)

I know the graph is a parabola. But I am not sure where I would "plug in" h and k into this equation. To get the translated equation.

1. You know

$\left|\begin{array}{l}\bar{x} = x+2 \\ \bar{y} =y-3\end{array}\right. ~\implies~ \left|\begin{array}{l}\bar{x}-2 = x \\ \bar{y}+3 =y\end{array}\right.$

2. Plug in the terms for x and y and change the variable:

$y+3 = 3(x-2)^2-2(x-2)+5$

Expand and collect like terms.
May 16th 2011, 03:33 PM
homeylova223

I am having difficulty with one more question

y^2+8x=0 rotated on pi/6

I did this

x= cos(pi/6)x+sin(pi/6)y
square root 3 / 2 x+1/2y

For y I got -1/2X+ square root 3/2y

Then I plug it in

(-1/2x+ square root 3/2y)^2+8( square root 3 /2 x +1/2y)

But I am having trouble figuring out the equation.
May 16th 2011, 10:13 PM
earboth

Quote:

Originally Posted by homeylova223

I am having difficulty with one more question

y^2+8x=0 rotated on pi/6

I did this

x= cos(pi/6)x+sin(pi/6)y
square root 3 / 2 x+1/2y

For y I got -1/2X+ square root 3/2y

Then I plug it in

(-1/2x+ square root 3/2y)^2+8( square root 3 /2 x +1/2y)

But I am having trouble figuring out the equation.

1. Do yourself (and us) a favour and start a new thread if you have a new question.

2. I assume that the center of rotation is the origin(?). If so you know:

$\left|\begin{array}{l}\bar{x} = x \cdot \cos\left(\frac{\pi}6\right) - y \cdot \sin\left(\frac{\pi}6\right) \\ \bar{y} = x \cdot \sin\left(\frac{\pi}6\right) + y \cdot \cos\left(\frac{\pi}6\right) \end{array}\right.$

With $\cos\left(\frac{\pi}6\right)=\frac12 \sqrt{3}$ and $\sin\left(\frac{\pi}6\right) = \frac12$

you'll get:

$\left|\begin{array}{l}x = \frac12 ( \bar{x} \sqrt{3} + \bar{y}) \\ y= \frac12 (\bar{y} \sqrt{3} - \bar{x} )\end{array}\right.$

3. Plug in these terms into the original equation which yields:

$\frac14 (\bar{y} \sqrt{3} - \bar{x})^2+ 4( \bar{x} \sqrt{3} + \bar{y}) = 0$

4. Expand the brackets and collect like terms:

$x^2 - 2·√3·x·y + 16·√3·x + 3·y^2 + 16·y = 0$
May 17th 2011, 10:29 AM
homeylova223

I have a question when you wrote the equation why is it square root of 3 instead of square root of 3 over 2? I mean it makes sense in the final answer.

Also would instead of having 1/2 in the equation would you not have 2 because if you plug it back into the equation with two it makes more sense to me

As I checked and found out the answer to this problem is

x^2-2square root 3 xy +3(y)^2+ 16square root 3 x+16y

And if you use 2 instead of 1/2 in the equation I would get that answer.
May 17th 2011, 10:48 PM
earboth

Quote:

Originally Posted by earboth

...
4. Expand the brackets and collect like terms:

$x^2 - 2·√3·x·y + 16·√3·x + 3·y^2 + 16·y = 0$

Quote:

Originally Posted by homeylova223

...
As I checked and found out the answer to this problem is

x^2-2square root 3 xy +3(y)^2+ 16square root 3 x+16y

And if you use 2 instead of 1/2 in the equation I would get that answer.

You probably have noticed that I made a mistake typing the equation (I used √3 instead of \sqrt{3}) So Here is the correct equation:

$x^2 - 2\cdot \sqrt{3} \cdot x·y + 16\cdot \sqrt{3} \cdot x + 3·y^2 + 16·y = 0$

Sorry for the confusion.