Are these answers regarding focus and directrix correct?












4












$begingroup$


1. Find the vertex, focus and directrix for the parabola given by



$y = 2x^2$



$(y-0) = 2(x-0)^2$



$(x-0)^2 = frac{1}{2}(y-0)$



$4p = frac{1}{2} implies p = frac{1}{8}$



V:$(0,0)$
F: $(0,frac{1}{8})$
D: $y = -frac{1}{8}$



2. Find the vertex, focus and directrix for the parabola given by
$$y = -2018x^2$$



$(x-0)^2 = -frac{1}{2018}(y-0)$



$4p = -frac{1}{2018} implies p = -frac{1}{8072}$



V: $(0,0)$
F: $(0,-frac{1}{8072})$
D: $y = frac{1}{8072}$



3. Find the vertex, focus and directrix for the parabola given by



$(y-2)^2 = 8(x+5)$



$4p = 8 implies p = 2$



V: $(-5,2)$
F: $(-3,2)$
D: $x= -7$



4. Find the vertex, focus and directrix for the parabola given by



$(y+6)^2 = frac{1}{2}(x-1)$



$4p = frac{1}{2} implies p = frac{1}{8}$



V:$(1,-6)$
F:$(frac{9}{8},-6)$
D:$x = frac{7}{8}$



5. Find the vertex, focus and directrix for the parabola given by



$y = 2x^2+5x-7$



$x^2+frac{5}{2}x-frac{7}{2} = frac{y}{2}$



$x^2+frac{5}{2}x = frac{y}{2} +frac{7}{2}$



$x^2+frac{5}{2}x +frac{25}{16}= frac{y}{2} +frac{7}{2}+frac{25}{16}$



$(x+frac{5}{4})^2 = frac{81}{16}+frac{y}{2}$



$(x+frac{5}{4})^2 = frac{1}{2}(frac{81}{8}+y)$



$4p = frac{1}{2} implies p = frac{1}{8}$



V:$(-frac{5}{2}, -frac{81}{8})$
F:$(-frac{5}{2}, -frac{80}{8})$
D:$y = -frac{82}{8}$



6. Find the vertex, focus and directrix for the parabola given by



$y = -frac{x^2}{4} - 2x +8$



$-4y = x^2 +8x -32$



$x^2 +8x +16 = 16+32+(-4y)$



$(x+4)^2 = 48 - 4y$



$(x+4)^2 = 4(12 - y)$



$(x+4)^2 = -4(y-12)$
$4p = -4 implies p= -1$



V:$(-4,12)$
F:$(-4,11)$
D:$y = 13$



7. Find the equation of the parabola which passes through the point $(8,12)$ with a vertex of $(4,-2)$



This one I'm a little confused about, aren't there multiple parabolas which could could through this point and have this vertex? I'm just going to assume it's squared in the x term, so:



$(x-4)^2 = 4p(y+2)$



$(8-4)^2 = 4p(12+2)$



$16 = 4p(14)$



$frac{16}{14} = 4p implies p = frac{4}{14}$



The equation is $(x-4)^2 = frac{16}{14} (y+2)$



8. Explain, with words, how the distance between the vertex and focus of a
parabola affects the steepness of the parabola



As the distance between the focus and directrix increases, $|p|$ decreases which means the parabola widens.






geometry conic-sections







share|cite|improve this question











$endgroup$

















    4












    $begingroup$


    1. Find the vertex, focus and directrix for the parabola given by



    $y = 2x^2$



    $(y-0) = 2(x-0)^2$



    $(x-0)^2 = frac{1}{2}(y-0)$



    $4p = frac{1}{2} implies p = frac{1}{8}$



    V:$(0,0)$
    F: $(0,frac{1}{8})$
    D: $y = -frac{1}{8}$



    2. Find the vertex, focus and directrix for the parabola given by
    $$y = -2018x^2$$



    $(x-0)^2 = -frac{1}{2018}(y-0)$



    $4p = -frac{1}{2018} implies p = -frac{1}{8072}$



    V: $(0,0)$
    F: $(0,-frac{1}{8072})$
    D: $y = frac{1}{8072}$



    3. Find the vertex, focus and directrix for the parabola given by



    $(y-2)^2 = 8(x+5)$



    $4p = 8 implies p = 2$



    V: $(-5,2)$
    F: $(-3,2)$
    D: $x= -7$



    4. Find the vertex, focus and directrix for the parabola given by



    $(y+6)^2 = frac{1}{2}(x-1)$



    $4p = frac{1}{2} implies p = frac{1}{8}$



    V:$(1,-6)$
    F:$(frac{9}{8},-6)$
    D:$x = frac{7}{8}$



    5. Find the vertex, focus and directrix for the parabola given by



    $y = 2x^2+5x-7$



    $x^2+frac{5}{2}x-frac{7}{2} = frac{y}{2}$



    $x^2+frac{5}{2}x = frac{y}{2} +frac{7}{2}$



    $x^2+frac{5}{2}x +frac{25}{16}= frac{y}{2} +frac{7}{2}+frac{25}{16}$



    $(x+frac{5}{4})^2 = frac{81}{16}+frac{y}{2}$



    $(x+frac{5}{4})^2 = frac{1}{2}(frac{81}{8}+y)$



    $4p = frac{1}{2} implies p = frac{1}{8}$



    V:$(-frac{5}{2}, -frac{81}{8})$
    F:$(-frac{5}{2}, -frac{80}{8})$
    D:$y = -frac{82}{8}$



    6. Find the vertex, focus and directrix for the parabola given by



    $y = -frac{x^2}{4} - 2x +8$



    $-4y = x^2 +8x -32$



    $x^2 +8x +16 = 16+32+(-4y)$



    $(x+4)^2 = 48 - 4y$



    $(x+4)^2 = 4(12 - y)$



    $(x+4)^2 = -4(y-12)$
    $4p = -4 implies p= -1$



    V:$(-4,12)$
    F:$(-4,11)$
    D:$y = 13$



    7. Find the equation of the parabola which passes through the point $(8,12)$ with a vertex of $(4,-2)$



    This one I'm a little confused about, aren't there multiple parabolas which could could through this point and have this vertex? I'm just going to assume it's squared in the x term, so:



    $(x-4)^2 = 4p(y+2)$



    $(8-4)^2 = 4p(12+2)$



    $16 = 4p(14)$



    $frac{16}{14} = 4p implies p = frac{4}{14}$



    The equation is $(x-4)^2 = frac{16}{14} (y+2)$



    8. Explain, with words, how the distance between the vertex and focus of a
    parabola affects the steepness of the parabola



    As the distance between the focus and directrix increases, $|p|$ decreases which means the parabola widens.






    geometry conic-sections







    share|cite|improve this question











    $endgroup$















      4












      4








      4





      $begingroup$


      1. Find the vertex, focus and directrix for the parabola given by



      $y = 2x^2$



      $(y-0) = 2(x-0)^2$



      $(x-0)^2 = frac{1}{2}(y-0)$



      $4p = frac{1}{2} implies p = frac{1}{8}$



      V:$(0,0)$
      F: $(0,frac{1}{8})$
      D: $y = -frac{1}{8}$



      2. Find the vertex, focus and directrix for the parabola given by
      $$y = -2018x^2$$



      $(x-0)^2 = -frac{1}{2018}(y-0)$



      $4p = -frac{1}{2018} implies p = -frac{1}{8072}$



      V: $(0,0)$
      F: $(0,-frac{1}{8072})$
      D: $y = frac{1}{8072}$



      3. Find the vertex, focus and directrix for the parabola given by



      $(y-2)^2 = 8(x+5)$



      $4p = 8 implies p = 2$



      V: $(-5,2)$
      F: $(-3,2)$
      D: $x= -7$



      4. Find the vertex, focus and directrix for the parabola given by



      $(y+6)^2 = frac{1}{2}(x-1)$



      $4p = frac{1}{2} implies p = frac{1}{8}$



      V:$(1,-6)$
      F:$(frac{9}{8},-6)$
      D:$x = frac{7}{8}$



      5. Find the vertex, focus and directrix for the parabola given by



      $y = 2x^2+5x-7$



      $x^2+frac{5}{2}x-frac{7}{2} = frac{y}{2}$



      $x^2+frac{5}{2}x = frac{y}{2} +frac{7}{2}$



      $x^2+frac{5}{2}x +frac{25}{16}= frac{y}{2} +frac{7}{2}+frac{25}{16}$



      $(x+frac{5}{4})^2 = frac{81}{16}+frac{y}{2}$



      $(x+frac{5}{4})^2 = frac{1}{2}(frac{81}{8}+y)$



      $4p = frac{1}{2} implies p = frac{1}{8}$



      V:$(-frac{5}{2}, -frac{81}{8})$
      F:$(-frac{5}{2}, -frac{80}{8})$
      D:$y = -frac{82}{8}$



      6. Find the vertex, focus and directrix for the parabola given by



      $y = -frac{x^2}{4} - 2x +8$



      $-4y = x^2 +8x -32$



      $x^2 +8x +16 = 16+32+(-4y)$



      $(x+4)^2 = 48 - 4y$



      $(x+4)^2 = 4(12 - y)$



      $(x+4)^2 = -4(y-12)$
      $4p = -4 implies p= -1$



      V:$(-4,12)$
      F:$(-4,11)$
      D:$y = 13$



      7. Find the equation of the parabola which passes through the point $(8,12)$ with a vertex of $(4,-2)$



      This one I'm a little confused about, aren't there multiple parabolas which could could through this point and have this vertex? I'm just going to assume it's squared in the x term, so:



      $(x-4)^2 = 4p(y+2)$



      $(8-4)^2 = 4p(12+2)$



      $16 = 4p(14)$



      $frac{16}{14} = 4p implies p = frac{4}{14}$



      The equation is $(x-4)^2 = frac{16}{14} (y+2)$



      8. Explain, with words, how the distance between the vertex and focus of a
      parabola affects the steepness of the parabola



      As the distance between the focus and directrix increases, $|p|$ decreases which means the parabola widens.






      geometry conic-sections







      share|cite|improve this question











      $endgroup$




      1. Find the vertex, focus and directrix for the parabola given by



      $y = 2x^2$



      $(y-0) = 2(x-0)^2$



      $(x-0)^2 = frac{1}{2}(y-0)$



      $4p = frac{1}{2} implies p = frac{1}{8}$



      V:$(0,0)$
      F: $(0,frac{1}{8})$
      D: $y = -frac{1}{8}$



      2. Find the vertex, focus and directrix for the parabola given by
      $$y = -2018x^2$$



      $(x-0)^2 = -frac{1}{2018}(y-0)$



      $4p = -frac{1}{2018} implies p = -frac{1}{8072}$



      V: $(0,0)$
      F: $(0,-frac{1}{8072})$
      D: $y = frac{1}{8072}$



      3. Find the vertex, focus and directrix for the parabola given by



      $(y-2)^2 = 8(x+5)$



      $4p = 8 implies p = 2$



      V: $(-5,2)$
      F: $(-3,2)$
      D: $x= -7$



      4. Find the vertex, focus and directrix for the parabola given by



      $(y+6)^2 = frac{1}{2}(x-1)$



      $4p = frac{1}{2} implies p = frac{1}{8}$



      V:$(1,-6)$
      F:$(frac{9}{8},-6)$
      D:$x = frac{7}{8}$



      5. Find the vertex, focus and directrix for the parabola given by



      $y = 2x^2+5x-7$



      $x^2+frac{5}{2}x-frac{7}{2} = frac{y}{2}$



      $x^2+frac{5}{2}x = frac{y}{2} +frac{7}{2}$



      $x^2+frac{5}{2}x +frac{25}{16}= frac{y}{2} +frac{7}{2}+frac{25}{16}$



      $(x+frac{5}{4})^2 = frac{81}{16}+frac{y}{2}$



      $(x+frac{5}{4})^2 = frac{1}{2}(frac{81}{8}+y)$



      $4p = frac{1}{2} implies p = frac{1}{8}$



      V:$(-frac{5}{2}, -frac{81}{8})$
      F:$(-frac{5}{2}, -frac{80}{8})$
      D:$y = -frac{82}{8}$



      6. Find the vertex, focus and directrix for the parabola given by



      $y = -frac{x^2}{4} - 2x +8$



      $-4y = x^2 +8x -32$



      $x^2 +8x +16 = 16+32+(-4y)$



      $(x+4)^2 = 48 - 4y$



      $(x+4)^2 = 4(12 - y)$



      $(x+4)^2 = -4(y-12)$
      $4p = -4 implies p= -1$



      V:$(-4,12)$
      F:$(-4,11)$
      D:$y = 13$



      7. Find the equation of the parabola which passes through the point $(8,12)$ with a vertex of $(4,-2)$



      This one I'm a little confused about, aren't there multiple parabolas which could could through this point and have this vertex? I'm just going to assume it's squared in the x term, so:



      $(x-4)^2 = 4p(y+2)$



      $(8-4)^2 = 4p(12+2)$



      $16 = 4p(14)$



      $frac{16}{14} = 4p implies p = frac{4}{14}$



      The equation is $(x-4)^2 = frac{16}{14} (y+2)$



      8. Explain, with words, how the distance between the vertex and focus of a
      parabola affects the steepness of the parabola



      As the distance between the focus and directrix increases, $|p|$ decreases which means the parabola widens.






      geometry conic-sections




      geometry conic-sections






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 25 '18 at 8:31







      user130306

















      asked Nov 25 '18 at 7:05









      user130306user130306

      40918




      40918






















          1 Answer
          1






          active

          oldest

          votes


















          0












          $begingroup$

          All the other answers are correct, equation of directrix is wrong in the fourth question (it should be $frac{7}{8}$), the final answer written in the seventh question is wrong $Big($putting value of 4p the final equation becomes $(x-4)^2 = frac{16}{14}(y+2)Big)$ and in the eighth question, we know that the distance between focus and directrix is 2|p|, therefore, as the distance increases, |p| increases.



          Hope it is helpful:)






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            thank you! can you explain how my answer for p is wrong in the 7th question?
            $endgroup$
            – user130306
            Nov 25 '18 at 8:30










          • $begingroup$
            Your answer for p is correct, but in the original equation you have to replace the value of 4p and not p only
            $endgroup$
            – Martund
            Nov 25 '18 at 8:32










          • $begingroup$
            also im confused about your last statement regarding question 8. i thought that as the distance ebtween focus and directrix increases, that |p| decreases
            $endgroup$
            – user130306
            Nov 25 '18 at 8:32










          • $begingroup$
            ohh i understand my issue for question 7, thank you
            $endgroup$
            – user130306
            Nov 25 '18 at 8:32










          • $begingroup$
            what is the distance between focus and directrix?
            $endgroup$
            – Martund
            Nov 25 '18 at 8:33











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          1 Answer
          1






          active

          oldest

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          active

          oldest

          votes






          active

          oldest

          votes









          0












          $begingroup$

          All the other answers are correct, equation of directrix is wrong in the fourth question (it should be $frac{7}{8}$), the final answer written in the seventh question is wrong $Big($putting value of 4p the final equation becomes $(x-4)^2 = frac{16}{14}(y+2)Big)$ and in the eighth question, we know that the distance between focus and directrix is 2|p|, therefore, as the distance increases, |p| increases.



          Hope it is helpful:)






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            thank you! can you explain how my answer for p is wrong in the 7th question?
            $endgroup$
            – user130306
            Nov 25 '18 at 8:30










          • $begingroup$
            Your answer for p is correct, but in the original equation you have to replace the value of 4p and not p only
            $endgroup$
            – Martund
            Nov 25 '18 at 8:32










          • $begingroup$
            also im confused about your last statement regarding question 8. i thought that as the distance ebtween focus and directrix increases, that |p| decreases
            $endgroup$
            – user130306
            Nov 25 '18 at 8:32










          • $begingroup$
            ohh i understand my issue for question 7, thank you
            $endgroup$
            – user130306
            Nov 25 '18 at 8:32










          • $begingroup$
            what is the distance between focus and directrix?
            $endgroup$
            – Martund
            Nov 25 '18 at 8:33
















          0












          $begingroup$

          All the other answers are correct, equation of directrix is wrong in the fourth question (it should be $frac{7}{8}$), the final answer written in the seventh question is wrong $Big($putting value of 4p the final equation becomes $(x-4)^2 = frac{16}{14}(y+2)Big)$ and in the eighth question, we know that the distance between focus and directrix is 2|p|, therefore, as the distance increases, |p| increases.



          Hope it is helpful:)






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            thank you! can you explain how my answer for p is wrong in the 7th question?
            $endgroup$
            – user130306
            Nov 25 '18 at 8:30










          • $begingroup$
            Your answer for p is correct, but in the original equation you have to replace the value of 4p and not p only
            $endgroup$
            – Martund
            Nov 25 '18 at 8:32










          • $begingroup$
            also im confused about your last statement regarding question 8. i thought that as the distance ebtween focus and directrix increases, that |p| decreases
            $endgroup$
            – user130306
            Nov 25 '18 at 8:32










          • $begingroup$
            ohh i understand my issue for question 7, thank you
            $endgroup$
            – user130306
            Nov 25 '18 at 8:32










          • $begingroup$
            what is the distance between focus and directrix?
            $endgroup$
            – Martund
            Nov 25 '18 at 8:33














          0












          0








          0





          $begingroup$

          All the other answers are correct, equation of directrix is wrong in the fourth question (it should be $frac{7}{8}$), the final answer written in the seventh question is wrong $Big($putting value of 4p the final equation becomes $(x-4)^2 = frac{16}{14}(y+2)Big)$ and in the eighth question, we know that the distance between focus and directrix is 2|p|, therefore, as the distance increases, |p| increases.



          Hope it is helpful:)






          share|cite|improve this answer









          $endgroup$



          All the other answers are correct, equation of directrix is wrong in the fourth question (it should be $frac{7}{8}$), the final answer written in the seventh question is wrong $Big($putting value of 4p the final equation becomes $(x-4)^2 = frac{16}{14}(y+2)Big)$ and in the eighth question, we know that the distance between focus and directrix is 2|p|, therefore, as the distance increases, |p| increases.



          Hope it is helpful:)







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 25 '18 at 8:23









          MartundMartund

          1,433212




          1,433212












          • $begingroup$
            thank you! can you explain how my answer for p is wrong in the 7th question?
            $endgroup$
            – user130306
            Nov 25 '18 at 8:30










          • $begingroup$
            Your answer for p is correct, but in the original equation you have to replace the value of 4p and not p only
            $endgroup$
            – Martund
            Nov 25 '18 at 8:32










          • $begingroup$
            also im confused about your last statement regarding question 8. i thought that as the distance ebtween focus and directrix increases, that |p| decreases
            $endgroup$
            – user130306
            Nov 25 '18 at 8:32










          • $begingroup$
            ohh i understand my issue for question 7, thank you
            $endgroup$
            – user130306
            Nov 25 '18 at 8:32










          • $begingroup$
            what is the distance between focus and directrix?
            $endgroup$
            – Martund
            Nov 25 '18 at 8:33


















          • $begingroup$
            thank you! can you explain how my answer for p is wrong in the 7th question?
            $endgroup$
            – user130306
            Nov 25 '18 at 8:30










          • $begingroup$
            Your answer for p is correct, but in the original equation you have to replace the value of 4p and not p only
            $endgroup$
            – Martund
            Nov 25 '18 at 8:32










          • $begingroup$
            also im confused about your last statement regarding question 8. i thought that as the distance ebtween focus and directrix increases, that |p| decreases
            $endgroup$
            – user130306
            Nov 25 '18 at 8:32










          • $begingroup$
            ohh i understand my issue for question 7, thank you
            $endgroup$
            – user130306
            Nov 25 '18 at 8:32










          • $begingroup$
            what is the distance between focus and directrix?
            $endgroup$
            – Martund
            Nov 25 '18 at 8:33
















          $begingroup$
          thank you! can you explain how my answer for p is wrong in the 7th question?
          $endgroup$
          – user130306
          Nov 25 '18 at 8:30




          $begingroup$
          thank you! can you explain how my answer for p is wrong in the 7th question?
          $endgroup$
          – user130306
          Nov 25 '18 at 8:30












          $begingroup$
          Your answer for p is correct, but in the original equation you have to replace the value of 4p and not p only
          $endgroup$
          – Martund
          Nov 25 '18 at 8:32




          $begingroup$
          Your answer for p is correct, but in the original equation you have to replace the value of 4p and not p only
          $endgroup$
          – Martund
          Nov 25 '18 at 8:32












          $begingroup$
          also im confused about your last statement regarding question 8. i thought that as the distance ebtween focus and directrix increases, that |p| decreases
          $endgroup$
          – user130306
          Nov 25 '18 at 8:32




          $begingroup$
          also im confused about your last statement regarding question 8. i thought that as the distance ebtween focus and directrix increases, that |p| decreases
          $endgroup$
          – user130306
          Nov 25 '18 at 8:32












          $begingroup$
          ohh i understand my issue for question 7, thank you
          $endgroup$
          – user130306
          Nov 25 '18 at 8:32




          $begingroup$
          ohh i understand my issue for question 7, thank you
          $endgroup$
          – user130306
          Nov 25 '18 at 8:32












          $begingroup$
          what is the distance between focus and directrix?
          $endgroup$
          – Martund
          Nov 25 '18 at 8:33




          $begingroup$
          what is the distance between focus and directrix?
          $endgroup$
          – Martund
          Nov 25 '18 at 8:33


















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