Converting parabola from Cartesian to parametric form
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I'm not quite sure how to convert this parabola equation in vertex form $y=a(x-h)^2 +k$ to parametric form. The equation is $y=-0.6x^2 +0.417$ and it has a focus at the origin and a vertical axis of symmetry.
I'm dying because this is due tomorrow please any help will be greatly appreciated.
algebra-precalculus conic-sections parametric curves
edited Aug 24 at 8:08
Rodrigo de Azevedo
12.7k41751
asked Aug 18 '16 at 8:51
Klowie Stewart
11
add a comment |Â
up vote
0
down vote
favorite
I'm not quite sure how to convert this parabola equation in vertex form $y=a(x-h)^2 +k$ to parametric form. The equation is $y=-0.6x^2 +0.417$ and it has a focus at the origin and a vertical axis of symmetry.
I'm dying because this is due tomorrow please any help will be greatly appreciated.
algebra-precalculus conic-sections parametric curves
edited Aug 24 at 8:08
Rodrigo de Azevedo
12.7k41751
asked Aug 18 '16 at 8:51
Klowie Stewart
11
Have you ever tried to google it?
– Ng Chung Tak
Aug 18 '16 at 13:46
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I'm not quite sure how to convert this parabola equation in vertex form $y=a(x-h)^2 +k$ to parametric form. The equation is $y=-0.6x^2 +0.417$ and it has a focus at the origin and a vertical axis of symmetry.
I'm dying because this is due tomorrow please any help will be greatly appreciated.
algebra-precalculus conic-sections parametric curves
edited Aug 24 at 8:08
Rodrigo de Azevedo
12.7k41751
asked Aug 18 '16 at 8:51
Klowie Stewart
11
I'm not quite sure how to convert this parabola equation in vertex form $y=a(x-h)^2 +k$ to parametric form. The equation is $y=-0.6x^2 +0.417$ and it has a focus at the origin and a vertical axis of symmetry.
I'm dying because this is due tomorrow please any help will be greatly appreciated.
algebra-precalculus conic-sections parametric curves
edited Aug 24 at 8:08
Rodrigo de Azevedo
12.7k41751
asked Aug 18 '16 at 8:51
Klowie Stewart
11
edited Aug 24 at 8:08
Rodrigo de Azevedo
12.7k41751
edited Aug 24 at 8:08
Rodrigo de Azevedo
12.7k41751
edited Aug 24 at 8:08
Rodrigo de Azevedo
12.7k41751
12.7k41751
asked Aug 18 '16 at 8:51
Klowie Stewart
11
asked Aug 18 '16 at 8:51
Klowie Stewart
11
asked Aug 18 '16 at 8:51
Klowie Stewart
11
11
Have you ever tried to google it?
– Ng Chung Tak
Aug 18 '16 at 13:46
add a comment |Â
Have you ever tried to google it?
– Ng Chung Tak
Aug 18 '16 at 13:46
Have you ever tried to google it?
– Ng Chung Tak
Aug 18 '16 at 13:46
Have you ever tried to google it?
– Ng Chung Tak
Aug 18 '16 at 13:46
add a comment |Â
1 Answer
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For the equation:
4a(y-k)=(x-h)^2,
the general formula for a parametric representation of this curve is:
x(t)=t+h+b and
y(t)=((t+b)^2)/(4a)+k
this is not the only general formula and the other one that I know of is closely related to this and I find this one slightly easier
In these equations, b is a 'real' scalar and changes the position on the curve where t=0.
The a in your equation is not the same as the a in the equations I have stated above. The a in mine is the focal length of the parabola. In fact (h,k) is the vertex of the parabola.
The simplest solution is to make b=0.
Wolfram has something on this but their parabola is sideways so x & y are swapped around.
http://mathworld.wolfram.com/Parabola.html
answered Nov 3 '16 at 0:00
john
1
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
-1
down vote
For the equation:
4a(y-k)=(x-h)^2,
the general formula for a parametric representation of this curve is:
x(t)=t+h+b and
y(t)=((t+b)^2)/(4a)+k
this is not the only general formula and the other one that I know of is closely related to this and I find this one slightly easier
In these equations, b is a 'real' scalar and changes the position on the curve where t=0.
The a in your equation is not the same as the a in the equations I have stated above. The a in mine is the focal length of the parabola. In fact (h,k) is the vertex of the parabola.
The simplest solution is to make b=0.
Wolfram has something on this but their parabola is sideways so x & y are swapped around.
http://mathworld.wolfram.com/Parabola.html
answered Nov 3 '16 at 0:00
john
1
add a comment |Â
up vote
-1
down vote
For the equation:
4a(y-k)=(x-h)^2,
the general formula for a parametric representation of this curve is:
x(t)=t+h+b and
y(t)=((t+b)^2)/(4a)+k
this is not the only general formula and the other one that I know of is closely related to this and I find this one slightly easier
In these equations, b is a 'real' scalar and changes the position on the curve where t=0.
The a in your equation is not the same as the a in the equations I have stated above. The a in mine is the focal length of the parabola. In fact (h,k) is the vertex of the parabola.
The simplest solution is to make b=0.
Wolfram has something on this but their parabola is sideways so x & y are swapped around.
http://mathworld.wolfram.com/Parabola.html
answered Nov 3 '16 at 0:00
john
1
add a comment |Â
up vote
-1
down vote
up vote
-1
down vote
For the equation:
4a(y-k)=(x-h)^2,
the general formula for a parametric representation of this curve is:
x(t)=t+h+b and
y(t)=((t+b)^2)/(4a)+k
this is not the only general formula and the other one that I know of is closely related to this and I find this one slightly easier
In these equations, b is a 'real' scalar and changes the position on the curve where t=0.
The a in your equation is not the same as the a in the equations I have stated above. The a in mine is the focal length of the parabola. In fact (h,k) is the vertex of the parabola.
The simplest solution is to make b=0.
Wolfram has something on this but their parabola is sideways so x & y are swapped around.
http://mathworld.wolfram.com/Parabola.html
answered Nov 3 '16 at 0:00
john
1
For the equation:
4a(y-k)=(x-h)^2,
the general formula for a parametric representation of this curve is:
x(t)=t+h+b and
y(t)=((t+b)^2)/(4a)+k
this is not the only general formula and the other one that I know of is closely related to this and I find this one slightly easier
In these equations, b is a 'real' scalar and changes the position on the curve where t=0.
The a in your equation is not the same as the a in the equations I have stated above. The a in mine is the focal length of the parabola. In fact (h,k) is the vertex of the parabola.
The simplest solution is to make b=0.
Wolfram has something on this but their parabola is sideways so x & y are swapped around.
http://mathworld.wolfram.com/Parabola.html
answered Nov 3 '16 at 0:00
john
1
answered Nov 3 '16 at 0:00
john
1
answered Nov 3 '16 at 0:00
john
1
answered Nov 3 '16 at 0:00
john
1
1
add a comment |Â
add a comment |Â
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Have you ever tried to google it?
– Ng Chung Tak
Aug 18 '16 at 13:46