Finding vertex of $xy^2=1$ .
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$$ xy^2 = 1$$
From the equation, it can be said that it is a rectangular hyperbola . Now if I want to calculate the vertex of it , I do follows :
$$rightarrow (fracx2+fracy^22)^2 - (fracx2 - frac y^22)^2=1$$
$$rightarrow X^2-Y^2=1$$
where, $X=fracx2+fracy^22$ and $Y=fracx2-fracy^22$ .
Now, for vertex :
$$X=pm a,Y=0 (a=1)$$
which implies :
$$ fracx2+fracy^22=pm 1 ....(i) $$
$$ fracx2-fracy^22=0 ....(ii) $$
Working with $(i)$ and $(ii)$ , I get :
$x=pm 2, y=0$ .
This, vertex is $(pm 2,0)$ .
Am I correct ?
conic-sections
asked Sep 1 at 9:30
Entrepreneur
339111
add a comment |Â
up vote
0
down vote
favorite
$$ xy^2 = 1$$
From the equation, it can be said that it is a rectangular hyperbola . Now if I want to calculate the vertex of it , I do follows :
$$rightarrow (fracx2+fracy^22)^2 - (fracx2 - frac y^22)^2=1$$
$$rightarrow X^2-Y^2=1$$
where, $X=fracx2+fracy^22$ and $Y=fracx2-fracy^22$ .
Now, for vertex :
$$X=pm a,Y=0 (a=1)$$
which implies :
$$ fracx2+fracy^22=pm 1 ....(i) $$
$$ fracx2-fracy^22=0 ....(ii) $$
Working with $(i)$ and $(ii)$ , I get :
$x=pm 2, y=0$ .
This, vertex is $(pm 2,0)$ .
Am I correct ?
conic-sections
asked Sep 1 at 9:30
Entrepreneur
339111
1
$xy^2=1$ is not a rectangular hyperbola, or any conic section. The equation of a conic has degree $2$, but $xy^2=1$ has degree $3$. Moreover, $(pm 2, 0)$ cannot be a vertex, since it does not satisfy the equation: $(pm 2)(0)^2 = 0 neq 1$.
– Blue
Sep 1 at 9:36
But the graph seems to be a hyperbola... Or is it only a curve like other 3-degree polynomials? @Blue
– Entrepreneur
Sep 1 at 9:43
2
Lots of graphs seem to be hyperbolas. :) This one, however, isn't terribly convincing: Its "branches" live in the first and fourth quadrants; a hyperbola's branches would be in opposite quadrants (say, first and third, or second and fourth). What you have is a cubic curve. These have been classified, but the classification is far more elaborate than ellipse/parabola/hyperbola.
– Blue
Sep 1 at 9:55
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
$$ xy^2 = 1$$
From the equation, it can be said that it is a rectangular hyperbola . Now if I want to calculate the vertex of it , I do follows :
$$rightarrow (fracx2+fracy^22)^2 - (fracx2 - frac y^22)^2=1$$
$$rightarrow X^2-Y^2=1$$
where, $X=fracx2+fracy^22$ and $Y=fracx2-fracy^22$ .
Now, for vertex :
$$X=pm a,Y=0 (a=1)$$
which implies :
$$ fracx2+fracy^22=pm 1 ....(i) $$
$$ fracx2-fracy^22=0 ....(ii) $$
Working with $(i)$ and $(ii)$ , I get :
$x=pm 2, y=0$ .
This, vertex is $(pm 2,0)$ .
Am I correct ?
conic-sections
asked Sep 1 at 9:30
Entrepreneur
339111
$$ xy^2 = 1$$
From the equation, it can be said that it is a rectangular hyperbola . Now if I want to calculate the vertex of it , I do follows :
$$rightarrow (fracx2+fracy^22)^2 - (fracx2 - frac y^22)^2=1$$
$$rightarrow X^2-Y^2=1$$
where, $X=fracx2+fracy^22$ and $Y=fracx2-fracy^22$ .
Now, for vertex :
$$X=pm a,Y=0 (a=1)$$
which implies :
$$ fracx2+fracy^22=pm 1 ....(i) $$
$$ fracx2-fracy^22=0 ....(ii) $$
Working with $(i)$ and $(ii)$ , I get :
$x=pm 2, y=0$ .
This, vertex is $(pm 2,0)$ .
Am I correct ?
conic-sections
conic-sections
asked Sep 1 at 9:30
Entrepreneur
339111
asked Sep 1 at 9:30
Entrepreneur
339111
asked Sep 1 at 9:30
Entrepreneur
339111
asked Sep 1 at 9:30
Entrepreneur
339111
asked Sep 1 at 9:30
Entrepreneur
339111
339111
1
$xy^2=1$ is not a rectangular hyperbola, or any conic section. The equation of a conic has degree $2$, but $xy^2=1$ has degree $3$. Moreover, $(pm 2, 0)$ cannot be a vertex, since it does not satisfy the equation: $(pm 2)(0)^2 = 0 neq 1$.
– Blue
Sep 1 at 9:36
But the graph seems to be a hyperbola... Or is it only a curve like other 3-degree polynomials? @Blue
– Entrepreneur
Sep 1 at 9:43
2
Lots of graphs seem to be hyperbolas. :) This one, however, isn't terribly convincing: Its "branches" live in the first and fourth quadrants; a hyperbola's branches would be in opposite quadrants (say, first and third, or second and fourth). What you have is a cubic curve. These have been classified, but the classification is far more elaborate than ellipse/parabola/hyperbola.
– Blue
Sep 1 at 9:55
add a comment |Â
1
$xy^2=1$ is not a rectangular hyperbola, or any conic section. The equation of a conic has degree $2$, but $xy^2=1$ has degree $3$. Moreover, $(pm 2, 0)$ cannot be a vertex, since it does not satisfy the equation: $(pm 2)(0)^2 = 0 neq 1$.
– Blue
Sep 1 at 9:36
But the graph seems to be a hyperbola... Or is it only a curve like other 3-degree polynomials? @Blue
– Entrepreneur
Sep 1 at 9:43
2
Lots of graphs seem to be hyperbolas. :) This one, however, isn't terribly convincing: Its "branches" live in the first and fourth quadrants; a hyperbola's branches would be in opposite quadrants (say, first and third, or second and fourth). What you have is a cubic curve. These have been classified, but the classification is far more elaborate than ellipse/parabola/hyperbola.
– Blue
Sep 1 at 9:55
1
1
$xy^2=1$ is not a rectangular hyperbola, or any conic section. The equation of a conic has degree $2$, but $xy^2=1$ has degree $3$. Moreover, $(pm 2, 0)$ cannot be a vertex, since it does not satisfy the equation: $(pm 2)(0)^2 = 0 neq 1$.
– Blue
Sep 1 at 9:36
$xy^2=1$ is not a rectangular hyperbola, or any conic section. The equation of a conic has degree $2$, but $xy^2=1$ has degree $3$. Moreover, $(pm 2, 0)$ cannot be a vertex, since it does not satisfy the equation: $(pm 2)(0)^2 = 0 neq 1$.
– Blue
Sep 1 at 9:36
But the graph seems to be a hyperbola... Or is it only a curve like other 3-degree polynomials? @Blue
– Entrepreneur
Sep 1 at 9:43
But the graph seems to be a hyperbola... Or is it only a curve like other 3-degree polynomials? @Blue
– Entrepreneur
Sep 1 at 9:43
2
2
Lots of graphs seem to be hyperbolas. :) This one, however, isn't terribly convincing: Its "branches" live in the first and fourth quadrants; a hyperbola's branches would be in opposite quadrants (say, first and third, or second and fourth). What you have is a cubic curve. These have been classified, but the classification is far more elaborate than ellipse/parabola/hyperbola.
– Blue
Sep 1 at 9:55
Lots of graphs seem to be hyperbolas. :) This one, however, isn't terribly convincing: Its "branches" live in the first and fourth quadrants; a hyperbola's branches would be in opposite quadrants (say, first and third, or second and fourth). What you have is a cubic curve. These have been classified, but the classification is far more elaborate than ellipse/parabola/hyperbola.
– Blue
Sep 1 at 9:55
add a comment |Â
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1
$xy^2=1$ is not a rectangular hyperbola, or any conic section. The equation of a conic has degree $2$, but $xy^2=1$ has degree $3$. Moreover, $(pm 2, 0)$ cannot be a vertex, since it does not satisfy the equation: $(pm 2)(0)^2 = 0 neq 1$.
– Blue
Sep 1 at 9:36
But the graph seems to be a hyperbola... Or is it only a curve like other 3-degree polynomials? @Blue
– Entrepreneur
Sep 1 at 9:43
2
Lots of graphs seem to be hyperbolas. :) This one, however, isn't terribly convincing: Its "branches" live in the first and fourth quadrants; a hyperbola's branches would be in opposite quadrants (say, first and third, or second and fourth). What you have is a cubic curve. These have been classified, but the classification is far more elaborate than ellipse/parabola/hyperbola.
– Blue
Sep 1 at 9:55