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 February 13th, 2013, 01:58 AM #1 Newbie   Joined: Feb 2013 Posts: 5 Thanks: 0 How To Plot This Graph? For the given below two equations: $x=(1-t^2)+x_{c}2t(1-t)+4t^2$ $y=(1-t^2)+y_{c}2t(1-t)+3t^2$ How to plot the graph if $x_{c}= 3$ and $y_{c}= 4$? $x=t^4 - 2t^2 +4t + 1$ $y=t^4 - 5t^2 + 6t + 1$
 February 13th, 2013, 02:41 AM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 522 Math Focus: Calculus/ODEs Re: How To Plot This Graph? With $x_c=3,\,y_c=4$ I get: $x(t)=1+6t-3t^2$ $y(t)=1+8t-6t^2$ and after eliminating the parameter, we have: $12x^2-12xy+3y^2-44x+30y+11=0$ which is a parabola. To sketch the graph, rotation of axes may be used.
February 13th, 2013, 06:54 PM   #3
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Re: How To Plot This Graph?

Quote:
 Originally Posted by MarkFL With $x_c=3,\,y_c=4$ I get: $x(t)=1+6t-3t^2$ $y(t)=1+8t-6t^2$ and after eliminating the parameter, we have: $12x^2-12xy+3y^2-44x+30y+11=0$ which is a parabola. To sketch the graph, rotation of axes may be used.
I think this is a wrong answer.

For the xc = 3 and yc = 4,I get:

$x=t^4 - 2t^2 +4t + 1$
$y=t^4 - 5t^2 + 6t + 1$

 February 13th, 2013, 06:56 PM #4 Newbie   Joined: Feb 2013 Posts: 5 Thanks: 0 Re: How To Plot This Graph? Oops, sorry so much. I am the wrong one.
February 13th, 2013, 06:57 PM   #5
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Re: How To Plot This Graph?

Quote:
 Originally Posted by MarkFL With $x_c=3,\,y_c=4$ I get: $x(t)=1+6t-3t^2$ $y(t)=1+8t-6t^2$ and after eliminating the parameter, we have: $12x^2-12xy+3y^2-44x+30y+11=0$ which is a parabola. To sketch the graph, rotation of axes may be used.
How do you eliminate the parameter?

How to?

 February 13th, 2013, 07:22 PM #6 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 522 Math Focus: Calculus/ODEs Re: How To Plot This Graph? What I did to eliminate the parameter is multiply the first parametric equation (for x) by -2, then add it to the other: $-2x=-2-12t+6t^2$ $y=1+8t-6t^2$ Adding these, we get: $y-2x=-1-4t$ and solving for $t$ we find: $t=\frac{2x-y-1}{4}$ Now, substitute for $t$ into either parametric equation. I chose the second: $y=1+8$$\frac{2x-y-1}{4}$$-6$$\frac{2x-y-1}{4}$$^2$ and this can be arranged as: $12x^2-12xy+3y^2-44x+30y+11=0$

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