(conic section) Find focus point given the equation of the parabola $ (3x^2-6x+2) $
$begingroup$
Judging by the picture, I don't think the focus point is correct, as it outside of the parabola.
I derived the coordinates of the parabola by first determining the a, b, and c value.
$$a=3, b=(-6), c=2$$
And then using the formula for the focus point:
$$F=(b/(2a), 1/(4a) + d/(4a))$$
Calculations
What is the correct focus point?
EDIT:
conic-sections
edited Jan 8 at 16:58
Key Flex
8,57761233
asked Jan 8 at 8:47
Ryan CameronRyan Cameron
697
$endgroup$
add a comment |
$begingroup$
Judging by the picture, I don't think the focus point is correct, as it outside of the parabola.
I derived the coordinates of the parabola by first determining the a, b, and c value.
$$a=3, b=(-6), c=2$$
And then using the formula for the focus point:
$$F=(b/(2a), 1/(4a) + d/(4a))$$
Calculations
What is the correct focus point?
EDIT:
conic-sections
edited Jan 8 at 16:58
Key Flex
8,57761233
asked Jan 8 at 8:47
Ryan CameronRyan Cameron
697
$endgroup$
1
$begingroup$
$x=-frac{b}{2a}$ same as the vertex
$endgroup$
– Jon
Jan 8 at 9:25
$begingroup$
Please elaborate.
$endgroup$
– Ryan Cameron
Jan 8 at 9:37
$begingroup$
This is a parabola which it's directrix is parallel to the $x$ axis, so the $x$ value at it's focus is the same as the one at the vertex.
$endgroup$
– Jon
Jan 8 at 9:45
$begingroup$
Where did you get this formula? It’s clearly missing a minus sign in the $x$-coordinate.
$endgroup$
– amd
Jan 8 at 20:29
add a comment |
$begingroup$
Judging by the picture, I don't think the focus point is correct, as it outside of the parabola.
I derived the coordinates of the parabola by first determining the a, b, and c value.
$$a=3, b=(-6), c=2$$
And then using the formula for the focus point:
$$F=(b/(2a), 1/(4a) + d/(4a))$$
Calculations
What is the correct focus point?
EDIT:
conic-sections
edited Jan 8 at 16:58
Key Flex
8,57761233
asked Jan 8 at 8:47
Ryan CameronRyan Cameron
697
$endgroup$
Judging by the picture, I don't think the focus point is correct, as it outside of the parabola.
I derived the coordinates of the parabola by first determining the a, b, and c value.
$$a=3, b=(-6), c=2$$
And then using the formula for the focus point:
$$F=(b/(2a), 1/(4a) + d/(4a))$$
Calculations
What is the correct focus point?
EDIT:
conic-sections
conic-sections
edited Jan 8 at 16:58
Key Flex
8,57761233
asked Jan 8 at 8:47
Ryan CameronRyan Cameron
697
edited Jan 8 at 16:58
Key Flex
8,57761233
asked Jan 8 at 8:47
Ryan CameronRyan Cameron
697
edited Jan 8 at 16:58
Key Flex
8,57761233
edited Jan 8 at 16:58
Key Flex
8,57761233
edited Jan 8 at 16:58
Key Flex
8,57761233
8,57761233
asked Jan 8 at 8:47
Ryan CameronRyan Cameron
697
asked Jan 8 at 8:47
Ryan CameronRyan Cameron
697
asked Jan 8 at 8:47
Ryan CameronRyan Cameron
697
697
1
$begingroup$
$x=-frac{b}{2a}$ same as the vertex
$endgroup$
– Jon
Jan 8 at 9:25
$begingroup$
Please elaborate.
$endgroup$
– Ryan Cameron
Jan 8 at 9:37
$begingroup$
This is a parabola which it's directrix is parallel to the $x$ axis, so the $x$ value at it's focus is the same as the one at the vertex.
$endgroup$
– Jon
Jan 8 at 9:45
$begingroup$
Where did you get this formula? It’s clearly missing a minus sign in the $x$-coordinate.
$endgroup$
– amd
Jan 8 at 20:29
add a comment |
1
$begingroup$
$x=-frac{b}{2a}$ same as the vertex
$endgroup$
– Jon
Jan 8 at 9:25
$begingroup$
Please elaborate.
$endgroup$
– Ryan Cameron
Jan 8 at 9:37
$begingroup$
This is a parabola which it's directrix is parallel to the $x$ axis, so the $x$ value at it's focus is the same as the one at the vertex.
$endgroup$
– Jon
Jan 8 at 9:45
$begingroup$
Where did you get this formula? It’s clearly missing a minus sign in the $x$-coordinate.
$endgroup$
– amd
Jan 8 at 20:29
1
1
$begingroup$
$x=-frac{b}{2a}$ same as the vertex
$endgroup$
– Jon
Jan 8 at 9:25
$begingroup$
$x=-frac{b}{2a}$ same as the vertex
$endgroup$
– Jon
Jan 8 at 9:25
$begingroup$
Please elaborate.
$endgroup$
– Ryan Cameron
Jan 8 at 9:37
$begingroup$
Please elaborate.
$endgroup$
– Ryan Cameron
Jan 8 at 9:37
$begingroup$
This is a parabola which it's directrix is parallel to the $x$ axis, so the $x$ value at it's focus is the same as the one at the vertex.
$endgroup$
– Jon
Jan 8 at 9:45
$begingroup$
This is a parabola which it's directrix is parallel to the $x$ axis, so the $x$ value at it's focus is the same as the one at the vertex.
$endgroup$
– Jon
Jan 8 at 9:45
$begingroup$
Where did you get this formula? It’s clearly missing a minus sign in the $x$-coordinate.
$endgroup$
– amd
Jan 8 at 20:29
$begingroup$
Where did you get this formula? It’s clearly missing a minus sign in the $x$-coordinate.
$endgroup$
– amd
Jan 8 at 20:29
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Better by differentiation with symbolic coefficients
$$ y= ax^2+bx+c;, y^{'}=2ax+b ;, y''=2a= 1/R $$ where $R$ is the radius of curvature, as $y^{'}=0$ in the curvature expression
The extremum point is
$$ (-b/2a, c-b^2/4a)$$ to whose $y$ coordinate we add $R/2=1/ 4a$ (sign automatically valid due to $a$ symbolically assigned ) to get $y$ coordinate of focal point as by virtue of a property of parabola its focal point bisects vertex-center of curvature line and this can be easily proved if we take parabola $ x^2= 4 f y.$ This gives us the formula for the focus.
$$ boxed{(-b/2a,, c+(1-b^2)/4a ) } $$
When $(a,b,c)=(3,-6,2)$ is plugged in, focal point coordinates are:
$$(1,-frac{11}{12}). $$
answered Jan 8 at 17:32
NarasimhamNarasimham
21.1k62258
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add a comment |
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1 Answer
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active
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votes
1 Answer
1
active
oldest
votes
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votes
active
oldest
votes
$begingroup$
Better by differentiation with symbolic coefficients
$$ y= ax^2+bx+c;, y^{'}=2ax+b ;, y''=2a= 1/R $$ where $R$ is the radius of curvature, as $y^{'}=0$ in the curvature expression
The extremum point is
$$ (-b/2a, c-b^2/4a)$$ to whose $y$ coordinate we add $R/2=1/ 4a$ (sign automatically valid due to $a$ symbolically assigned ) to get $y$ coordinate of focal point as by virtue of a property of parabola its focal point bisects vertex-center of curvature line and this can be easily proved if we take parabola $ x^2= 4 f y.$ This gives us the formula for the focus.
$$ boxed{(-b/2a,, c+(1-b^2)/4a ) } $$
When $(a,b,c)=(3,-6,2)$ is plugged in, focal point coordinates are:
$$(1,-frac{11}{12}). $$
answered Jan 8 at 17:32
NarasimhamNarasimham
21.1k62258
$endgroup$
add a comment |
$begingroup$
Better by differentiation with symbolic coefficients
$$ y= ax^2+bx+c;, y^{'}=2ax+b ;, y''=2a= 1/R $$ where $R$ is the radius of curvature, as $y^{'}=0$ in the curvature expression
The extremum point is
$$ (-b/2a, c-b^2/4a)$$ to whose $y$ coordinate we add $R/2=1/ 4a$ (sign automatically valid due to $a$ symbolically assigned ) to get $y$ coordinate of focal point as by virtue of a property of parabola its focal point bisects vertex-center of curvature line and this can be easily proved if we take parabola $ x^2= 4 f y.$ This gives us the formula for the focus.
$$ boxed{(-b/2a,, c+(1-b^2)/4a ) } $$
When $(a,b,c)=(3,-6,2)$ is plugged in, focal point coordinates are:
$$(1,-frac{11}{12}). $$
answered Jan 8 at 17:32
NarasimhamNarasimham
21.1k62258
$endgroup$
add a comment |
$begingroup$
Better by differentiation with symbolic coefficients
$$ y= ax^2+bx+c;, y^{'}=2ax+b ;, y''=2a= 1/R $$ where $R$ is the radius of curvature, as $y^{'}=0$ in the curvature expression
The extremum point is
$$ (-b/2a, c-b^2/4a)$$ to whose $y$ coordinate we add $R/2=1/ 4a$ (sign automatically valid due to $a$ symbolically assigned ) to get $y$ coordinate of focal point as by virtue of a property of parabola its focal point bisects vertex-center of curvature line and this can be easily proved if we take parabola $ x^2= 4 f y.$ This gives us the formula for the focus.
$$ boxed{(-b/2a,, c+(1-b^2)/4a ) } $$
When $(a,b,c)=(3,-6,2)$ is plugged in, focal point coordinates are:
$$(1,-frac{11}{12}). $$
answered Jan 8 at 17:32
NarasimhamNarasimham
21.1k62258
$endgroup$
Better by differentiation with symbolic coefficients
$$ y= ax^2+bx+c;, y^{'}=2ax+b ;, y''=2a= 1/R $$ where $R$ is the radius of curvature, as $y^{'}=0$ in the curvature expression
The extremum point is
$$ (-b/2a, c-b^2/4a)$$ to whose $y$ coordinate we add $R/2=1/ 4a$ (sign automatically valid due to $a$ symbolically assigned ) to get $y$ coordinate of focal point as by virtue of a property of parabola its focal point bisects vertex-center of curvature line and this can be easily proved if we take parabola $ x^2= 4 f y.$ This gives us the formula for the focus.
$$ boxed{(-b/2a,, c+(1-b^2)/4a ) } $$
When $(a,b,c)=(3,-6,2)$ is plugged in, focal point coordinates are:
$$(1,-frac{11}{12}). $$
answered Jan 8 at 17:32
NarasimhamNarasimham
21.1k62258
edited Jan 9 at 14:20
answered Jan 8 at 17:32
NarasimhamNarasimham
21.1k62258
answered Jan 8 at 17:32
NarasimhamNarasimham
21.1k62258
answered Jan 8 at 17:32
NarasimhamNarasimham
21.1k62258
21.1k62258
add a comment |
add a comment |
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1
$begingroup$
$x=-frac{b}{2a}$ same as the vertex
$endgroup$
– Jon
Jan 8 at 9:25
$begingroup$
Please elaborate.
$endgroup$
– Ryan Cameron
Jan 8 at 9:37
$begingroup$
This is a parabola which it's directrix is parallel to the $x$ axis, so the $x$ value at it's focus is the same as the one at the vertex.
$endgroup$
– Jon
Jan 8 at 9:45
$begingroup$
Where did you get this formula? It’s clearly missing a minus sign in the $x$-coordinate.
$endgroup$
– amd
Jan 8 at 20:29