Curve Tracing: Why I'm getting asymptotes parallel to both X & Y Axis?

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











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0
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Here is equation of curve.
$ x^2 y^2=a ^2 (y ^2- x^2) $



Here is my calculations...tell me if something is wrong in it.
My Calculations




calculus geometry curves




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  • Because they both exist.
    – Michael Hoppe
    Aug 3 at 8:18










  • How can that exist??
    – user580502
    Aug 3 at 8:18










  • Both on x & y axis??
    – user580502
    Aug 3 at 8:18










  • We can't even draw such curve...
    – user580502
    Aug 3 at 8:19






  • 1




    You should write out your calculations in MathJaX. As a sidenote the equality $-a^2 = y^2$ has no real solutions.
    – Triatticus
    Aug 3 at 8:20















up vote
0
down vote

favorite












Here is equation of curve.
$ x^2 y^2=a ^2 (y ^2- x^2) $



Here is my calculations...tell me if something is wrong in it.
My Calculations




calculus geometry curves




share|cite|improve this question



















  • Because they both exist.
    – Michael Hoppe
    Aug 3 at 8:18










  • How can that exist??
    – user580502
    Aug 3 at 8:18










  • Both on x & y axis??
    – user580502
    Aug 3 at 8:18










  • We can't even draw such curve...
    – user580502
    Aug 3 at 8:19






  • 1




    You should write out your calculations in MathJaX. As a sidenote the equality $-a^2 = y^2$ has no real solutions.
    – Triatticus
    Aug 3 at 8:20













up vote
0
down vote

favorite









up vote
0
down vote

favorite











Here is equation of curve.
$ x^2 y^2=a ^2 (y ^2- x^2) $



Here is my calculations...tell me if something is wrong in it.
My Calculations




calculus geometry curves




share|cite|improve this question











Here is equation of curve.
$ x^2 y^2=a ^2 (y ^2- x^2) $



Here is my calculations...tell me if something is wrong in it.
My Calculations





calculus geometry curves





share|cite|improve this question










share|cite|improve this question




share|cite|improve this question









asked Aug 3 at 8:14









user580502

83




83











  • Because they both exist.
    – Michael Hoppe
    Aug 3 at 8:18










  • How can that exist??
    – user580502
    Aug 3 at 8:18










  • Both on x & y axis??
    – user580502
    Aug 3 at 8:18










  • We can't even draw such curve...
    – user580502
    Aug 3 at 8:19






  • 1




    You should write out your calculations in MathJaX. As a sidenote the equality $-a^2 = y^2$ has no real solutions.
    – Triatticus
    Aug 3 at 8:20

















  • Because they both exist.
    – Michael Hoppe
    Aug 3 at 8:18










  • How can that exist??
    – user580502
    Aug 3 at 8:18










  • Both on x & y axis??
    – user580502
    Aug 3 at 8:18










  • We can't even draw such curve...
    – user580502
    Aug 3 at 8:19






  • 1




    You should write out your calculations in MathJaX. As a sidenote the equality $-a^2 = y^2$ has no real solutions.
    – Triatticus
    Aug 3 at 8:20
















Because they both exist.
– Michael Hoppe
Aug 3 at 8:18




Because they both exist.
– Michael Hoppe
Aug 3 at 8:18












How can that exist??
– user580502
Aug 3 at 8:18




How can that exist??
– user580502
Aug 3 at 8:18












Both on x & y axis??
– user580502
Aug 3 at 8:18




Both on x & y axis??
– user580502
Aug 3 at 8:18












We can't even draw such curve...
– user580502
Aug 3 at 8:19




We can't even draw such curve...
– user580502
Aug 3 at 8:19




1




1




You should write out your calculations in MathJaX. As a sidenote the equality $-a^2 = y^2$ has no real solutions.
– Triatticus
Aug 3 at 8:20





You should write out your calculations in MathJaX. As a sidenote the equality $-a^2 = y^2$ has no real solutions.
– Triatticus
Aug 3 at 8:20











1 Answer
1






active

oldest

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up vote
0
down vote



accepted










Solving for $x^2$ and $y^2$ we have



$$
x^2 = fraca^2 y^2a^2+ y^2\
y^2 = fraca^2 x^2a^2-x^2
$$



In the first, we have



$$
lim_ytoinfty x^2 = a^2
$$



and in the second



$$
lim_x_pmto pm a = pminfty
$$



so the asymptotes are located at $x = pm a$



enter image description here






share|cite|improve this answer





















  • Thank You soo much !!!!!!
    – user580502
    Aug 3 at 9:16










Your Answer




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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
0
down vote



accepted










Solving for $x^2$ and $y^2$ we have



$$
x^2 = fraca^2 y^2a^2+ y^2\
y^2 = fraca^2 x^2a^2-x^2
$$



In the first, we have



$$
lim_ytoinfty x^2 = a^2
$$



and in the second



$$
lim_x_pmto pm a = pminfty
$$



so the asymptotes are located at $x = pm a$



enter image description here






share|cite|improve this answer





















  • Thank You soo much !!!!!!
    – user580502
    Aug 3 at 9:16














up vote
0
down vote



accepted










Solving for $x^2$ and $y^2$ we have



$$
x^2 = fraca^2 y^2a^2+ y^2\
y^2 = fraca^2 x^2a^2-x^2
$$



In the first, we have



$$
lim_ytoinfty x^2 = a^2
$$



and in the second



$$
lim_x_pmto pm a = pminfty
$$



so the asymptotes are located at $x = pm a$



enter image description here






share|cite|improve this answer





















  • Thank You soo much !!!!!!
    – user580502
    Aug 3 at 9:16












up vote
0
down vote



accepted







up vote
0
down vote



accepted






Solving for $x^2$ and $y^2$ we have



$$
x^2 = fraca^2 y^2a^2+ y^2\
y^2 = fraca^2 x^2a^2-x^2
$$



In the first, we have



$$
lim_ytoinfty x^2 = a^2
$$



and in the second



$$
lim_x_pmto pm a = pminfty
$$



so the asymptotes are located at $x = pm a$



enter image description here






share|cite|improve this answer













Solving for $x^2$ and $y^2$ we have



$$
x^2 = fraca^2 y^2a^2+ y^2\
y^2 = fraca^2 x^2a^2-x^2
$$



In the first, we have



$$
lim_ytoinfty x^2 = a^2
$$



and in the second



$$
lim_x_pmto pm a = pminfty
$$



so the asymptotes are located at $x = pm a$



enter image description here







share|cite|improve this answer













share|cite|improve this answer



share|cite|improve this answer











answered Aug 3 at 8:47









Cesareo

5,4912412




5,4912412











  • Thank You soo much !!!!!!
    – user580502
    Aug 3 at 9:16
















  • Thank You soo much !!!!!!
    – user580502
    Aug 3 at 9:16















Thank You soo much !!!!!!
– user580502
Aug 3 at 9:16




Thank You soo much !!!!!!
– user580502
Aug 3 at 9:16












 

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