Find shortest distance from the parabola $y=x^2-9$ to the origin.












2












$begingroup$



Find shortest distance from the parabola $y=x^2-9$ to the origin.




First, I find minima of $sqrt{x^2+(x^2-9)^2}$, so use derivative and ...



Is have an easier way?






optimization conic-sections maxima-minima







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    With distance, it's usually easier to drop the square root. After all, minimizing the distance is the same as minimizing the squared distance.
    $endgroup$
    – lulu
    Jan 12 at 13:26
















2












$begingroup$



Find shortest distance from the parabola $y=x^2-9$ to the origin.




First, I find minima of $sqrt{x^2+(x^2-9)^2}$, so use derivative and ...



Is have an easier way?






optimization conic-sections maxima-minima







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    With distance, it's usually easier to drop the square root. After all, minimizing the distance is the same as minimizing the squared distance.
    $endgroup$
    – lulu
    Jan 12 at 13:26














2












2








2


1



$begingroup$



Find shortest distance from the parabola $y=x^2-9$ to the origin.




First, I find minima of $sqrt{x^2+(x^2-9)^2}$, so use derivative and ...



Is have an easier way?






optimization conic-sections maxima-minima







share|cite|improve this question











$endgroup$





Find shortest distance from the parabola $y=x^2-9$ to the origin.




First, I find minima of $sqrt{x^2+(x^2-9)^2}$, so use derivative and ...



Is have an easier way?






optimization conic-sections maxima-minima




optimization conic-sections maxima-minima






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 12 at 14:35









Michael Rozenberg

103k1891195




103k1891195










asked Jan 12 at 13:22









HeartHeart

28418




28418








  • 2




    $begingroup$
    With distance, it's usually easier to drop the square root. After all, minimizing the distance is the same as minimizing the squared distance.
    $endgroup$
    – lulu
    Jan 12 at 13:26














  • 2




    $begingroup$
    With distance, it's usually easier to drop the square root. After all, minimizing the distance is the same as minimizing the squared distance.
    $endgroup$
    – lulu
    Jan 12 at 13:26








2




2




$begingroup$
With distance, it's usually easier to drop the square root. After all, minimizing the distance is the same as minimizing the squared distance.
$endgroup$
– lulu
Jan 12 at 13:26




$begingroup$
With distance, it's usually easier to drop the square root. After all, minimizing the distance is the same as minimizing the squared distance.
$endgroup$
– lulu
Jan 12 at 13:26










6 Answers
6






active

oldest

votes


















4












$begingroup$

Another approach: if $P=(x,y)$ is the point of minimum on the parabola, then line $OP$ must be perpendicular to the tangent at $P$. Hence:
$$2x=-{xover y},quadtext{that is}quad y=-{1over2}.$$






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    Here is an easier way: the shortest distance $r$ is taken at the minimum of
    $r^2 = y^2 + x^2 = y^2 + y + 9$ This gets minimal at $y = -0.5$ (take the derivative or write $r^2 = (y+ 0.5)^2 + 8.75 ge 8.75 $), so $r = sqrt{y^2 + y + 9} ge sqrt{8.75} simeq 2.9580$.






    share|cite|improve this answer











    $endgroup$





















      1












      $begingroup$

      $$x^2+(x^2-9)^2=(x^2-9)^2+x^2-9+frac{1}{4}+frac{35}{4}=left(x^2-9+frac{1}{2}right)^2+frac{35}{4}geqfrac{35}{4}.$$
      The equality occurs for $x^2=frac{17}{2},$ which says that $frac{sqrt{35}}{2}$ is a minimal value.






      share|cite|improve this answer









      $endgroup$





















        1












        $begingroup$

        The direct way is the easier way.
        $$ d^2 = x^2+ (x^2-9)^2 = x^4 -17x^2+81 $$
        Differentiating $d^2$ (since derivative point is same as with $d$) and removing common facor $2x$
        $$ 2 x^2-17 =0,, x_{min}=sqrt{17/2}, y_{min}= -1/2, d_{min}=sqrt{x_{min}^2+y_{min}^2} = sqrt{35/4}$$
        One more differentiation can verify minimum here.






        share|cite|improve this answer











        $endgroup$





















          0












          $begingroup$

          The distance $d$ is minimized when $d^2$ is minimized. We have $d^2=(x^2-9)^2+x^2. $ Substituting $x^2=z$ we have $d^2=(z-9)^2+z=z^2-17z+81.......$ I will assume you can find the minimum value of a quadratic in $z$.






          share|cite|improve this answer









          $endgroup$





















            0












            $begingroup$

            firstly, we have:
            $$f(x)=x^2-9=(x+3)(x-3)$$
            from this we can see that $f(-x)=f(x)$. At $x=3$ and $x=-3$ the distance from the origin is $3$ so $l_{min}le3$ and this can be our upper limit.
            If you want the exact answer differentiation is the best way to do it. Note:
            $$l=sqrt{x^2+(x^2-9)^2}$$
            $$l^2=x^2+(x^2-9)^2$$
            $$2lfrac{dl}{dx}=2x+4x(x^2-9)$$
            It is a bit of a shortcut for the differentiation






            share|cite|improve this answer









            $endgroup$













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              6 Answers
              6






              active

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              6 Answers
              6






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              active

              oldest

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              active

              oldest

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              4












              $begingroup$

              Another approach: if $P=(x,y)$ is the point of minimum on the parabola, then line $OP$ must be perpendicular to the tangent at $P$. Hence:
              $$2x=-{xover y},quadtext{that is}quad y=-{1over2}.$$






              share|cite|improve this answer









              $endgroup$


















                4












                $begingroup$

                Another approach: if $P=(x,y)$ is the point of minimum on the parabola, then line $OP$ must be perpendicular to the tangent at $P$. Hence:
                $$2x=-{xover y},quadtext{that is}quad y=-{1over2}.$$






                share|cite|improve this answer









                $endgroup$
















                  4












                  4








                  4





                  $begingroup$

                  Another approach: if $P=(x,y)$ is the point of minimum on the parabola, then line $OP$ must be perpendicular to the tangent at $P$. Hence:
                  $$2x=-{xover y},quadtext{that is}quad y=-{1over2}.$$






                  share|cite|improve this answer









                  $endgroup$



                  Another approach: if $P=(x,y)$ is the point of minimum on the parabola, then line $OP$ must be perpendicular to the tangent at $P$. Hence:
                  $$2x=-{xover y},quadtext{that is}quad y=-{1over2}.$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 12 at 13:40









                  AretinoAretino

                  23.6k21443




                  23.6k21443























                      2












                      $begingroup$

                      Here is an easier way: the shortest distance $r$ is taken at the minimum of
                      $r^2 = y^2 + x^2 = y^2 + y + 9$ This gets minimal at $y = -0.5$ (take the derivative or write $r^2 = (y+ 0.5)^2 + 8.75 ge 8.75 $), so $r = sqrt{y^2 + y + 9} ge sqrt{8.75} simeq 2.9580$.






                      share|cite|improve this answer











                      $endgroup$


















                        2












                        $begingroup$

                        Here is an easier way: the shortest distance $r$ is taken at the minimum of
                        $r^2 = y^2 + x^2 = y^2 + y + 9$ This gets minimal at $y = -0.5$ (take the derivative or write $r^2 = (y+ 0.5)^2 + 8.75 ge 8.75 $), so $r = sqrt{y^2 + y + 9} ge sqrt{8.75} simeq 2.9580$.






                        share|cite|improve this answer











                        $endgroup$
















                          2












                          2








                          2





                          $begingroup$

                          Here is an easier way: the shortest distance $r$ is taken at the minimum of
                          $r^2 = y^2 + x^2 = y^2 + y + 9$ This gets minimal at $y = -0.5$ (take the derivative or write $r^2 = (y+ 0.5)^2 + 8.75 ge 8.75 $), so $r = sqrt{y^2 + y + 9} ge sqrt{8.75} simeq 2.9580$.






                          share|cite|improve this answer











                          $endgroup$



                          Here is an easier way: the shortest distance $r$ is taken at the minimum of
                          $r^2 = y^2 + x^2 = y^2 + y + 9$ This gets minimal at $y = -0.5$ (take the derivative or write $r^2 = (y+ 0.5)^2 + 8.75 ge 8.75 $), so $r = sqrt{y^2 + y + 9} ge sqrt{8.75} simeq 2.9580$.







                          share|cite|improve this answer














                          share|cite|improve this answer



                          share|cite|improve this answer








                          edited Jan 12 at 13:34

























                          answered Jan 12 at 13:29









                          AndreasAndreas

                          8,1131037




                          8,1131037























                              1












                              $begingroup$

                              $$x^2+(x^2-9)^2=(x^2-9)^2+x^2-9+frac{1}{4}+frac{35}{4}=left(x^2-9+frac{1}{2}right)^2+frac{35}{4}geqfrac{35}{4}.$$
                              The equality occurs for $x^2=frac{17}{2},$ which says that $frac{sqrt{35}}{2}$ is a minimal value.






                              share|cite|improve this answer









                              $endgroup$


















                                1












                                $begingroup$

                                $$x^2+(x^2-9)^2=(x^2-9)^2+x^2-9+frac{1}{4}+frac{35}{4}=left(x^2-9+frac{1}{2}right)^2+frac{35}{4}geqfrac{35}{4}.$$
                                The equality occurs for $x^2=frac{17}{2},$ which says that $frac{sqrt{35}}{2}$ is a minimal value.






                                share|cite|improve this answer









                                $endgroup$
















                                  1












                                  1








                                  1





                                  $begingroup$

                                  $$x^2+(x^2-9)^2=(x^2-9)^2+x^2-9+frac{1}{4}+frac{35}{4}=left(x^2-9+frac{1}{2}right)^2+frac{35}{4}geqfrac{35}{4}.$$
                                  The equality occurs for $x^2=frac{17}{2},$ which says that $frac{sqrt{35}}{2}$ is a minimal value.






                                  share|cite|improve this answer









                                  $endgroup$



                                  $$x^2+(x^2-9)^2=(x^2-9)^2+x^2-9+frac{1}{4}+frac{35}{4}=left(x^2-9+frac{1}{2}right)^2+frac{35}{4}geqfrac{35}{4}.$$
                                  The equality occurs for $x^2=frac{17}{2},$ which says that $frac{sqrt{35}}{2}$ is a minimal value.







                                  share|cite|improve this answer












                                  share|cite|improve this answer



                                  share|cite|improve this answer










                                  answered Jan 12 at 14:17









                                  Michael RozenbergMichael Rozenberg

                                  103k1891195




                                  103k1891195























                                      1












                                      $begingroup$

                                      The direct way is the easier way.
                                      $$ d^2 = x^2+ (x^2-9)^2 = x^4 -17x^2+81 $$
                                      Differentiating $d^2$ (since derivative point is same as with $d$) and removing common facor $2x$
                                      $$ 2 x^2-17 =0,, x_{min}=sqrt{17/2}, y_{min}= -1/2, d_{min}=sqrt{x_{min}^2+y_{min}^2} = sqrt{35/4}$$
                                      One more differentiation can verify minimum here.






                                      share|cite|improve this answer











                                      $endgroup$


















                                        1












                                        $begingroup$

                                        The direct way is the easier way.
                                        $$ d^2 = x^2+ (x^2-9)^2 = x^4 -17x^2+81 $$
                                        Differentiating $d^2$ (since derivative point is same as with $d$) and removing common facor $2x$
                                        $$ 2 x^2-17 =0,, x_{min}=sqrt{17/2}, y_{min}= -1/2, d_{min}=sqrt{x_{min}^2+y_{min}^2} = sqrt{35/4}$$
                                        One more differentiation can verify minimum here.






                                        share|cite|improve this answer











                                        $endgroup$
















                                          1












                                          1








                                          1





                                          $begingroup$

                                          The direct way is the easier way.
                                          $$ d^2 = x^2+ (x^2-9)^2 = x^4 -17x^2+81 $$
                                          Differentiating $d^2$ (since derivative point is same as with $d$) and removing common facor $2x$
                                          $$ 2 x^2-17 =0,, x_{min}=sqrt{17/2}, y_{min}= -1/2, d_{min}=sqrt{x_{min}^2+y_{min}^2} = sqrt{35/4}$$
                                          One more differentiation can verify minimum here.






                                          share|cite|improve this answer











                                          $endgroup$



                                          The direct way is the easier way.
                                          $$ d^2 = x^2+ (x^2-9)^2 = x^4 -17x^2+81 $$
                                          Differentiating $d^2$ (since derivative point is same as with $d$) and removing common facor $2x$
                                          $$ 2 x^2-17 =0,, x_{min}=sqrt{17/2}, y_{min}= -1/2, d_{min}=sqrt{x_{min}^2+y_{min}^2} = sqrt{35/4}$$
                                          One more differentiation can verify minimum here.







                                          share|cite|improve this answer














                                          share|cite|improve this answer



                                          share|cite|improve this answer








                                          edited Jan 12 at 15:31

























                                          answered Jan 12 at 14:59









                                          NarasimhamNarasimham

                                          20.8k52158




                                          20.8k52158























                                              0












                                              $begingroup$

                                              The distance $d$ is minimized when $d^2$ is minimized. We have $d^2=(x^2-9)^2+x^2. $ Substituting $x^2=z$ we have $d^2=(z-9)^2+z=z^2-17z+81.......$ I will assume you can find the minimum value of a quadratic in $z$.






                                              share|cite|improve this answer









                                              $endgroup$


















                                                0












                                                $begingroup$

                                                The distance $d$ is minimized when $d^2$ is minimized. We have $d^2=(x^2-9)^2+x^2. $ Substituting $x^2=z$ we have $d^2=(z-9)^2+z=z^2-17z+81.......$ I will assume you can find the minimum value of a quadratic in $z$.






                                                share|cite|improve this answer









                                                $endgroup$
















                                                  0












                                                  0








                                                  0





                                                  $begingroup$

                                                  The distance $d$ is minimized when $d^2$ is minimized. We have $d^2=(x^2-9)^2+x^2. $ Substituting $x^2=z$ we have $d^2=(z-9)^2+z=z^2-17z+81.......$ I will assume you can find the minimum value of a quadratic in $z$.






                                                  share|cite|improve this answer









                                                  $endgroup$



                                                  The distance $d$ is minimized when $d^2$ is minimized. We have $d^2=(x^2-9)^2+x^2. $ Substituting $x^2=z$ we have $d^2=(z-9)^2+z=z^2-17z+81.......$ I will assume you can find the minimum value of a quadratic in $z$.







                                                  share|cite|improve this answer












                                                  share|cite|improve this answer



                                                  share|cite|improve this answer










                                                  answered Jan 12 at 13:42









                                                  DanielWainfleetDanielWainfleet

                                                  35k31648




                                                  35k31648























                                                      0












                                                      $begingroup$

                                                      firstly, we have:
                                                      $$f(x)=x^2-9=(x+3)(x-3)$$
                                                      from this we can see that $f(-x)=f(x)$. At $x=3$ and $x=-3$ the distance from the origin is $3$ so $l_{min}le3$ and this can be our upper limit.
                                                      If you want the exact answer differentiation is the best way to do it. Note:
                                                      $$l=sqrt{x^2+(x^2-9)^2}$$
                                                      $$l^2=x^2+(x^2-9)^2$$
                                                      $$2lfrac{dl}{dx}=2x+4x(x^2-9)$$
                                                      It is a bit of a shortcut for the differentiation






                                                      share|cite|improve this answer









                                                      $endgroup$


















                                                        0












                                                        $begingroup$

                                                        firstly, we have:
                                                        $$f(x)=x^2-9=(x+3)(x-3)$$
                                                        from this we can see that $f(-x)=f(x)$. At $x=3$ and $x=-3$ the distance from the origin is $3$ so $l_{min}le3$ and this can be our upper limit.
                                                        If you want the exact answer differentiation is the best way to do it. Note:
                                                        $$l=sqrt{x^2+(x^2-9)^2}$$
                                                        $$l^2=x^2+(x^2-9)^2$$
                                                        $$2lfrac{dl}{dx}=2x+4x(x^2-9)$$
                                                        It is a bit of a shortcut for the differentiation






                                                        share|cite|improve this answer









                                                        $endgroup$
















                                                          0












                                                          0








                                                          0





                                                          $begingroup$

                                                          firstly, we have:
                                                          $$f(x)=x^2-9=(x+3)(x-3)$$
                                                          from this we can see that $f(-x)=f(x)$. At $x=3$ and $x=-3$ the distance from the origin is $3$ so $l_{min}le3$ and this can be our upper limit.
                                                          If you want the exact answer differentiation is the best way to do it. Note:
                                                          $$l=sqrt{x^2+(x^2-9)^2}$$
                                                          $$l^2=x^2+(x^2-9)^2$$
                                                          $$2lfrac{dl}{dx}=2x+4x(x^2-9)$$
                                                          It is a bit of a shortcut for the differentiation






                                                          share|cite|improve this answer









                                                          $endgroup$



                                                          firstly, we have:
                                                          $$f(x)=x^2-9=(x+3)(x-3)$$
                                                          from this we can see that $f(-x)=f(x)$. At $x=3$ and $x=-3$ the distance from the origin is $3$ so $l_{min}le3$ and this can be our upper limit.
                                                          If you want the exact answer differentiation is the best way to do it. Note:
                                                          $$l=sqrt{x^2+(x^2-9)^2}$$
                                                          $$l^2=x^2+(x^2-9)^2$$
                                                          $$2lfrac{dl}{dx}=2x+4x(x^2-9)$$
                                                          It is a bit of a shortcut for the differentiation







                                                          share|cite|improve this answer












                                                          share|cite|improve this answer



                                                          share|cite|improve this answer










                                                          answered Jan 12 at 13:43









                                                          Henry LeeHenry Lee

                                                          1,966219




                                                          1,966219






























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