How to plot $(x^2+2xy-24)^2+(2x^2+y^2-33)^2=0$ by hand?
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I am trying to solve this problem from Kindle's Analytic Geometry book (Chapter 2, problem 11). I have to plot, by hand, the equation:
$(x^2+2xy-24)^2+(2x^2+y^2-33)^2=0$
I can't figure out whether it is a rotated conic section or something like that. I wouldn't like to expand the expression since I am supposed to plot it in a short time for a quick quiz we'll have in class.
geometry analytic-geometry graphing-functions
asked Dec 1 '18 at 16:07
MSalmerMSalmer
32
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I am trying to solve this problem from Kindle's Analytic Geometry book (Chapter 2, problem 11). I have to plot, by hand, the equation:
$(x^2+2xy-24)^2+(2x^2+y^2-33)^2=0$
I can't figure out whether it is a rotated conic section or something like that. I wouldn't like to expand the expression since I am supposed to plot it in a short time for a quick quiz we'll have in class.
geometry analytic-geometry graphing-functions
asked Dec 1 '18 at 16:07
MSalmerMSalmer
32
$endgroup$
add a comment |
$begingroup$
I am trying to solve this problem from Kindle's Analytic Geometry book (Chapter 2, problem 11). I have to plot, by hand, the equation:
$(x^2+2xy-24)^2+(2x^2+y^2-33)^2=0$
I can't figure out whether it is a rotated conic section or something like that. I wouldn't like to expand the expression since I am supposed to plot it in a short time for a quick quiz we'll have in class.
geometry analytic-geometry graphing-functions
asked Dec 1 '18 at 16:07
MSalmerMSalmer
32
$endgroup$
I am trying to solve this problem from Kindle's Analytic Geometry book (Chapter 2, problem 11). I have to plot, by hand, the equation:
$(x^2+2xy-24)^2+(2x^2+y^2-33)^2=0$
I can't figure out whether it is a rotated conic section or something like that. I wouldn't like to expand the expression since I am supposed to plot it in a short time for a quick quiz we'll have in class.
geometry analytic-geometry graphing-functions
geometry analytic-geometry graphing-functions
asked Dec 1 '18 at 16:07
MSalmerMSalmer
32
asked Dec 1 '18 at 16:07
MSalmerMSalmer
32
asked Dec 1 '18 at 16:07
MSalmerMSalmer
32
asked Dec 1 '18 at 16:07
MSalmerMSalmer
32
asked Dec 1 '18 at 16:07
MSalmerMSalmer
32
32
add a comment |
add a comment |
1 Answer
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Apparently, the task requires to find the coordinates of the points that satisfy the equation, which holds true when:
$$begin{cases}x^2+2xy-24=0 \ 2x^2+y^2-33=0end{cases} Rightarrow begin{cases}y=frac{24-x^2}{2x} \ 2x^2+left(frac{24-x^2}{2x}right)^2-33=0end{cases} Rightarrow x^4-20x^2+64=0 Rightarrow \
x_{1,2,3,4}=pm 2,pm 4 Rightarrow y_{1,2,3,4}=5,-5,1,-1.$$
So, you put the four points on the plane and you are done.
answered Dec 1 '18 at 16:57
farruhotafarruhota
19.7k2738
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1 Answer
1
active
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1 Answer
1
active
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active
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$begingroup$
Apparently, the task requires to find the coordinates of the points that satisfy the equation, which holds true when:
$$begin{cases}x^2+2xy-24=0 \ 2x^2+y^2-33=0end{cases} Rightarrow begin{cases}y=frac{24-x^2}{2x} \ 2x^2+left(frac{24-x^2}{2x}right)^2-33=0end{cases} Rightarrow x^4-20x^2+64=0 Rightarrow \
x_{1,2,3,4}=pm 2,pm 4 Rightarrow y_{1,2,3,4}=5,-5,1,-1.$$
So, you put the four points on the plane and you are done.
answered Dec 1 '18 at 16:57
farruhotafarruhota
19.7k2738
$endgroup$
add a comment |
$begingroup$
Apparently, the task requires to find the coordinates of the points that satisfy the equation, which holds true when:
$$begin{cases}x^2+2xy-24=0 \ 2x^2+y^2-33=0end{cases} Rightarrow begin{cases}y=frac{24-x^2}{2x} \ 2x^2+left(frac{24-x^2}{2x}right)^2-33=0end{cases} Rightarrow x^4-20x^2+64=0 Rightarrow \
x_{1,2,3,4}=pm 2,pm 4 Rightarrow y_{1,2,3,4}=5,-5,1,-1.$$
So, you put the four points on the plane and you are done.
answered Dec 1 '18 at 16:57
farruhotafarruhota
19.7k2738
$endgroup$
add a comment |
$begingroup$
Apparently, the task requires to find the coordinates of the points that satisfy the equation, which holds true when:
$$begin{cases}x^2+2xy-24=0 \ 2x^2+y^2-33=0end{cases} Rightarrow begin{cases}y=frac{24-x^2}{2x} \ 2x^2+left(frac{24-x^2}{2x}right)^2-33=0end{cases} Rightarrow x^4-20x^2+64=0 Rightarrow \
x_{1,2,3,4}=pm 2,pm 4 Rightarrow y_{1,2,3,4}=5,-5,1,-1.$$
So, you put the four points on the plane and you are done.
answered Dec 1 '18 at 16:57
farruhotafarruhota
19.7k2738
$endgroup$
Apparently, the task requires to find the coordinates of the points that satisfy the equation, which holds true when:
$$begin{cases}x^2+2xy-24=0 \ 2x^2+y^2-33=0end{cases} Rightarrow begin{cases}y=frac{24-x^2}{2x} \ 2x^2+left(frac{24-x^2}{2x}right)^2-33=0end{cases} Rightarrow x^4-20x^2+64=0 Rightarrow \
x_{1,2,3,4}=pm 2,pm 4 Rightarrow y_{1,2,3,4}=5,-5,1,-1.$$
So, you put the four points on the plane and you are done.
answered Dec 1 '18 at 16:57
farruhotafarruhota
19.7k2738
answered Dec 1 '18 at 16:57
farruhotafarruhota
19.7k2738
answered Dec 1 '18 at 16:57
farruhotafarruhota
19.7k2738
answered Dec 1 '18 at 16:57
farruhotafarruhota
19.7k2738
19.7k2738
add a comment |
add a comment |
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