(conic section) Find focus point given the equation of the parabola $ (3x^2-6x+2) $












2












$begingroup$


enter image description here



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



enter image description here




What is the correct focus point?




EDIT:



enter image description here






conic-sections







share|cite|improve this question











$endgroup$








  • 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


















2












$begingroup$


enter image description here



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



enter image description here




What is the correct focus point?




EDIT:



enter image description here






conic-sections







share|cite|improve this question











$endgroup$








  • 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
















2












2








2





$begingroup$


enter image description here



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



enter image description here




What is the correct focus point?




EDIT:



enter image description here






conic-sections







share|cite|improve this question











$endgroup$




enter image description here



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



enter image description here




What is the correct focus point?




EDIT:



enter image description here






conic-sections




conic-sections






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 8 at 16:58









Key Flex

8,56171233




8,56171233










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
















  • 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












1 Answer
1






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oldest

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1












$begingroup$

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:



$$(1,-frac{11}{12}). $$






share|cite|improve this answer











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    1 Answer
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    active

    oldest

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    $begingroup$

    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:



    $$(1,-frac{11}{12}). $$






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      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:



      $$(1,-frac{11}{12}). $$






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        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:



        $$(1,-frac{11}{12}). $$






        share|cite|improve this answer











        $endgroup$



        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:



        $$(1,-frac{11}{12}). $$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 9 at 14:20

























        answered Jan 8 at 17:32









        NarasimhamNarasimham

        21.2k62258




        21.2k62258






























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