How to find the triangle area inside the parabola? Please help me understand.












0












$begingroup$


The parabola $C$ has cartesian equation $y^2 = 12x.$

The point $P(3p^2, 6p)$ lies on $C,$ where $pneq0.$




  • (a) Show that the equation of the normal to the curve $C$ at the point $P$ is
    $$y + px = 6p + 3p^3$$
    This normal crosses the curve $C$ again at the point $Q.$

    Given that $p = 2$ and that $S$ is the focus of the parabola, find


  • (b) the coordinates of the point $Q,$


  • (c) the area of the triangle $PQS.$



I can't figure out a way to solve question (c). I know the answer but don't understand it.






analytic-geometry conic-sections







share|cite|improve this question











$endgroup$












  • $begingroup$
    Are you familiar with: $text{Area}=frac12 big| (x_P - x_Q) (y_S - y_P) - (x_P - x_S) (y_Q - y_P) big|$? If not, see en.wikipedia.org/wiki/Shoelace_formula
    $endgroup$
    – Anurag A
    Dec 26 '18 at 20:10








  • 1




    $begingroup$
    Have you solved (a) and (b)? How did you do that? What did you get?
    $endgroup$
    – Arthur
    Dec 26 '18 at 20:11










  • $begingroup$
    Thanks @AnuragA, Such an intricate formula, no wonder I couldn't' solve it for the life of me.
    $endgroup$
    – Antonio
    Dec 26 '18 at 20:29










  • $begingroup$
    If $R$ is the point where line $PQ$ crosses the x-axis, then it is easy to compute the areas of triangles $PRS$ and $QRS$.
    $endgroup$
    – Aretino
    Dec 28 '18 at 15:52
















0












$begingroup$


The parabola $C$ has cartesian equation $y^2 = 12x.$

The point $P(3p^2, 6p)$ lies on $C,$ where $pneq0.$




  • (a) Show that the equation of the normal to the curve $C$ at the point $P$ is
    $$y + px = 6p + 3p^3$$
    This normal crosses the curve $C$ again at the point $Q.$

    Given that $p = 2$ and that $S$ is the focus of the parabola, find


  • (b) the coordinates of the point $Q,$


  • (c) the area of the triangle $PQS.$



I can't figure out a way to solve question (c). I know the answer but don't understand it.






analytic-geometry conic-sections







share|cite|improve this question











$endgroup$












  • $begingroup$
    Are you familiar with: $text{Area}=frac12 big| (x_P - x_Q) (y_S - y_P) - (x_P - x_S) (y_Q - y_P) big|$? If not, see en.wikipedia.org/wiki/Shoelace_formula
    $endgroup$
    – Anurag A
    Dec 26 '18 at 20:10








  • 1




    $begingroup$
    Have you solved (a) and (b)? How did you do that? What did you get?
    $endgroup$
    – Arthur
    Dec 26 '18 at 20:11










  • $begingroup$
    Thanks @AnuragA, Such an intricate formula, no wonder I couldn't' solve it for the life of me.
    $endgroup$
    – Antonio
    Dec 26 '18 at 20:29










  • $begingroup$
    If $R$ is the point where line $PQ$ crosses the x-axis, then it is easy to compute the areas of triangles $PRS$ and $QRS$.
    $endgroup$
    – Aretino
    Dec 28 '18 at 15:52














0












0








0





$begingroup$


The parabola $C$ has cartesian equation $y^2 = 12x.$

The point $P(3p^2, 6p)$ lies on $C,$ where $pneq0.$




  • (a) Show that the equation of the normal to the curve $C$ at the point $P$ is
    $$y + px = 6p + 3p^3$$
    This normal crosses the curve $C$ again at the point $Q.$

    Given that $p = 2$ and that $S$ is the focus of the parabola, find


  • (b) the coordinates of the point $Q,$


  • (c) the area of the triangle $PQS.$



I can't figure out a way to solve question (c). I know the answer but don't understand it.






analytic-geometry conic-sections







share|cite|improve this question











$endgroup$




The parabola $C$ has cartesian equation $y^2 = 12x.$

The point $P(3p^2, 6p)$ lies on $C,$ where $pneq0.$




  • (a) Show that the equation of the normal to the curve $C$ at the point $P$ is
    $$y + px = 6p + 3p^3$$
    This normal crosses the curve $C$ again at the point $Q.$

    Given that $p = 2$ and that $S$ is the focus of the parabola, find


  • (b) the coordinates of the point $Q,$


  • (c) the area of the triangle $PQS.$



I can't figure out a way to solve question (c). I know the answer but don't understand it.






analytic-geometry conic-sections




analytic-geometry conic-sections






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 31 '18 at 11:41









user376343

3,9584829




3,9584829










asked Dec 26 '18 at 20:07









AntonioAntonio

11




11












  • $begingroup$
    Are you familiar with: $text{Area}=frac12 big| (x_P - x_Q) (y_S - y_P) - (x_P - x_S) (y_Q - y_P) big|$? If not, see en.wikipedia.org/wiki/Shoelace_formula
    $endgroup$
    – Anurag A
    Dec 26 '18 at 20:10








  • 1




    $begingroup$
    Have you solved (a) and (b)? How did you do that? What did you get?
    $endgroup$
    – Arthur
    Dec 26 '18 at 20:11










  • $begingroup$
    Thanks @AnuragA, Such an intricate formula, no wonder I couldn't' solve it for the life of me.
    $endgroup$
    – Antonio
    Dec 26 '18 at 20:29










  • $begingroup$
    If $R$ is the point where line $PQ$ crosses the x-axis, then it is easy to compute the areas of triangles $PRS$ and $QRS$.
    $endgroup$
    – Aretino
    Dec 28 '18 at 15:52


















  • $begingroup$
    Are you familiar with: $text{Area}=frac12 big| (x_P - x_Q) (y_S - y_P) - (x_P - x_S) (y_Q - y_P) big|$? If not, see en.wikipedia.org/wiki/Shoelace_formula
    $endgroup$
    – Anurag A
    Dec 26 '18 at 20:10








  • 1




    $begingroup$
    Have you solved (a) and (b)? How did you do that? What did you get?
    $endgroup$
    – Arthur
    Dec 26 '18 at 20:11










  • $begingroup$
    Thanks @AnuragA, Such an intricate formula, no wonder I couldn't' solve it for the life of me.
    $endgroup$
    – Antonio
    Dec 26 '18 at 20:29










  • $begingroup$
    If $R$ is the point where line $PQ$ crosses the x-axis, then it is easy to compute the areas of triangles $PRS$ and $QRS$.
    $endgroup$
    – Aretino
    Dec 28 '18 at 15:52
















$begingroup$
Are you familiar with: $text{Area}=frac12 big| (x_P - x_Q) (y_S - y_P) - (x_P - x_S) (y_Q - y_P) big|$? If not, see en.wikipedia.org/wiki/Shoelace_formula
$endgroup$
– Anurag A
Dec 26 '18 at 20:10






$begingroup$
Are you familiar with: $text{Area}=frac12 big| (x_P - x_Q) (y_S - y_P) - (x_P - x_S) (y_Q - y_P) big|$? If not, see en.wikipedia.org/wiki/Shoelace_formula
$endgroup$
– Anurag A
Dec 26 '18 at 20:10






1




1




$begingroup$
Have you solved (a) and (b)? How did you do that? What did you get?
$endgroup$
– Arthur
Dec 26 '18 at 20:11




$begingroup$
Have you solved (a) and (b)? How did you do that? What did you get?
$endgroup$
– Arthur
Dec 26 '18 at 20:11












$begingroup$
Thanks @AnuragA, Such an intricate formula, no wonder I couldn't' solve it for the life of me.
$endgroup$
– Antonio
Dec 26 '18 at 20:29




$begingroup$
Thanks @AnuragA, Such an intricate formula, no wonder I couldn't' solve it for the life of me.
$endgroup$
– Antonio
Dec 26 '18 at 20:29












$begingroup$
If $R$ is the point where line $PQ$ crosses the x-axis, then it is easy to compute the areas of triangles $PRS$ and $QRS$.
$endgroup$
– Aretino
Dec 28 '18 at 15:52




$begingroup$
If $R$ is the point where line $PQ$ crosses the x-axis, then it is easy to compute the areas of triangles $PRS$ and $QRS$.
$endgroup$
– Aretino
Dec 28 '18 at 15:52










2 Answers
2






active

oldest

votes


















1












$begingroup$

The slope of the tangent line in $$P(3p^2,6p)$$ is given by $$y'=frac{6}{y}$$ so our nomal line has the slope $$y'_N=-p$$ and the equation is given by
$$y=-px+n$$ and plug in the coordinates of $P$ we get
$$6p=-3p^3+n$$ thus $$y=-px+6p+3p^3$$ is the equation of our normal line.
And für $p=2$ you have to solve
$$(-2x+36)^2=12x$$ for $$x$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    And what about the triangle area PQS?
    $endgroup$
    – Antonio
    Dec 26 '18 at 20:32



















0












$begingroup$

For $p=2$ we have: $P=(12,12)$, $Q=(27,-18)$, $S=(3,0)$. Moreover, line $PQ$ intersects the $x$-axis at $R=(18,0)$. It follows that triangles $PRS$ and $QRS$ have base $RS=15$ in common and altitudes $P_y=12$, $|Q_y|=18$, so that:
$$
area_{PSQ}=area_{PRS}+area_{QRS}={1over2}15cdot12+{1over2}15cdot18=225.
$$






share|cite|improve this answer









$endgroup$













    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3053268%2fhow-to-find-the-triangle-area-inside-the-parabola-please-help-me-understand%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    The slope of the tangent line in $$P(3p^2,6p)$$ is given by $$y'=frac{6}{y}$$ so our nomal line has the slope $$y'_N=-p$$ and the equation is given by
    $$y=-px+n$$ and plug in the coordinates of $P$ we get
    $$6p=-3p^3+n$$ thus $$y=-px+6p+3p^3$$ is the equation of our normal line.
    And für $p=2$ you have to solve
    $$(-2x+36)^2=12x$$ for $$x$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      And what about the triangle area PQS?
      $endgroup$
      – Antonio
      Dec 26 '18 at 20:32
















    1












    $begingroup$

    The slope of the tangent line in $$P(3p^2,6p)$$ is given by $$y'=frac{6}{y}$$ so our nomal line has the slope $$y'_N=-p$$ and the equation is given by
    $$y=-px+n$$ and plug in the coordinates of $P$ we get
    $$6p=-3p^3+n$$ thus $$y=-px+6p+3p^3$$ is the equation of our normal line.
    And für $p=2$ you have to solve
    $$(-2x+36)^2=12x$$ for $$x$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      And what about the triangle area PQS?
      $endgroup$
      – Antonio
      Dec 26 '18 at 20:32














    1












    1








    1





    $begingroup$

    The slope of the tangent line in $$P(3p^2,6p)$$ is given by $$y'=frac{6}{y}$$ so our nomal line has the slope $$y'_N=-p$$ and the equation is given by
    $$y=-px+n$$ and plug in the coordinates of $P$ we get
    $$6p=-3p^3+n$$ thus $$y=-px+6p+3p^3$$ is the equation of our normal line.
    And für $p=2$ you have to solve
    $$(-2x+36)^2=12x$$ for $$x$$






    share|cite|improve this answer









    $endgroup$



    The slope of the tangent line in $$P(3p^2,6p)$$ is given by $$y'=frac{6}{y}$$ so our nomal line has the slope $$y'_N=-p$$ and the equation is given by
    $$y=-px+n$$ and plug in the coordinates of $P$ we get
    $$6p=-3p^3+n$$ thus $$y=-px+6p+3p^3$$ is the equation of our normal line.
    And für $p=2$ you have to solve
    $$(-2x+36)^2=12x$$ for $$x$$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 26 '18 at 20:17









    Dr. Sonnhard GraubnerDr. Sonnhard Graubner

    77.9k42866




    77.9k42866












    • $begingroup$
      And what about the triangle area PQS?
      $endgroup$
      – Antonio
      Dec 26 '18 at 20:32


















    • $begingroup$
      And what about the triangle area PQS?
      $endgroup$
      – Antonio
      Dec 26 '18 at 20:32
















    $begingroup$
    And what about the triangle area PQS?
    $endgroup$
    – Antonio
    Dec 26 '18 at 20:32




    $begingroup$
    And what about the triangle area PQS?
    $endgroup$
    – Antonio
    Dec 26 '18 at 20:32











    0












    $begingroup$

    For $p=2$ we have: $P=(12,12)$, $Q=(27,-18)$, $S=(3,0)$. Moreover, line $PQ$ intersects the $x$-axis at $R=(18,0)$. It follows that triangles $PRS$ and $QRS$ have base $RS=15$ in common and altitudes $P_y=12$, $|Q_y|=18$, so that:
    $$
    area_{PSQ}=area_{PRS}+area_{QRS}={1over2}15cdot12+{1over2}15cdot18=225.
    $$






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      For $p=2$ we have: $P=(12,12)$, $Q=(27,-18)$, $S=(3,0)$. Moreover, line $PQ$ intersects the $x$-axis at $R=(18,0)$. It follows that triangles $PRS$ and $QRS$ have base $RS=15$ in common and altitudes $P_y=12$, $|Q_y|=18$, so that:
      $$
      area_{PSQ}=area_{PRS}+area_{QRS}={1over2}15cdot12+{1over2}15cdot18=225.
      $$






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        For $p=2$ we have: $P=(12,12)$, $Q=(27,-18)$, $S=(3,0)$. Moreover, line $PQ$ intersects the $x$-axis at $R=(18,0)$. It follows that triangles $PRS$ and $QRS$ have base $RS=15$ in common and altitudes $P_y=12$, $|Q_y|=18$, so that:
        $$
        area_{PSQ}=area_{PRS}+area_{QRS}={1over2}15cdot12+{1over2}15cdot18=225.
        $$






        share|cite|improve this answer









        $endgroup$



        For $p=2$ we have: $P=(12,12)$, $Q=(27,-18)$, $S=(3,0)$. Moreover, line $PQ$ intersects the $x$-axis at $R=(18,0)$. It follows that triangles $PRS$ and $QRS$ have base $RS=15$ in common and altitudes $P_y=12$, $|Q_y|=18$, so that:
        $$
        area_{PSQ}=area_{PRS}+area_{QRS}={1over2}15cdot12+{1over2}15cdot18=225.
        $$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 29 '18 at 16:10









        AretinoAretino

        25.2k21445




        25.2k21445






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3053268%2fhow-to-find-the-triangle-area-inside-the-parabola-please-help-me-understand%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            Probability when a professor distributes a quiz and homework assignment to a class of n students.

            Image
            0 0 $begingroup$ Need help with this problem. Suppose our lazy professor collects a quiz and a homework assignment from a class of n students one day, then distributes both the quizzes and the homework assignments back to the class in a random fashion for grading. Each student receives one quiz and one homework assignment to grade. (a) What is the probability that every student receives someone else's quiz to grade, and someone else's homework to grade? (b) What is the probability that no student receives both their own quiz and their own homework assignment to grade? In this case, some students may receive their own quiz, and others may receive their own homework assignment. (c) Compute the limiting probability as n approaches infinity in each case. ...
            Read more

            Aardman Animations

            Image
            Aardman Animations, Ltd. Aardman Animations Tipo privada Atividade Animação em stop-motion , Animação em CGI Fundação 1972 Fundador(es) Peter Lord David Sproxton Sede Bristol, Inglaterra  Reino Unido Proprietário(s) Dreamworks Pessoas-chave Peter Lord David Sproxton Nick Park Divisões Aardman Features Aardman Digital Aardman Commercials Aardman Broadcast Aardman International Aardman Rights Aardman Effects Aardman 3-D Systems Aardman Nathan Love AardBoiled Website oficial aardman.com ...
            Read more

            Are they similar matrix

            Image
            0 $begingroup$ Do $begin{bmatrix} 0&i&0\0&0&1\0&0&0 end{bmatrix} $ and $begin{bmatrix} 0&0&0\-i&0&0\0&1&0 end{bmatrix} $ are similar.Is this True/false Clearly both are nilpotent and one is conjucate transpose of other but how to know if they are similar.i'm stuck. Please help me linear-algebra vector-spaces eigenvalues-eigenvectors generalizedeigenvector share | cite | improve this question edited Dec 10 '18 at 6:38 Vasanth Kris asked Dec 10 '18 at 6:19 ...
            Read more