Are these answers regarding focus and directrix correct?
$begingroup$
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
asked Nov 25 '18 at 7:05
user130306user130306
40918
$endgroup$
add a comment |
$begingroup$
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
asked Nov 25 '18 at 7:05
user130306user130306
40918
$endgroup$
add a comment |
$begingroup$
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
asked Nov 25 '18 at 7:05
user130306user130306
40918
$endgroup$
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
asked Nov 25 '18 at 7:05
user130306user130306
40918
asked Nov 25 '18 at 7:05
user130306user130306
40918
edited Nov 25 '18 at 8:31
user130306
asked Nov 25 '18 at 7:05
user130306user130306
40918
asked Nov 25 '18 at 7:05
user130306user130306
40918
asked Nov 25 '18 at 7:05
user130306user130306
40918
40918
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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:)
answered Nov 25 '18 at 8:23
MartundMartund
1,433212
$endgroup$
$begingroup$
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
|
show 5 more comments
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$begingroup$
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:)
answered Nov 25 '18 at 8:23
MartundMartund
1,433212
$endgroup$
$begingroup$
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
|
show 5 more comments
$begingroup$
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:)
answered Nov 25 '18 at 8:23
MartundMartund
1,433212
$endgroup$
$begingroup$
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
|
show 5 more comments
$begingroup$
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:)
answered Nov 25 '18 at 8:23
MartundMartund
1,433212
$endgroup$
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:)
answered Nov 25 '18 at 8:23
MartundMartund
1,433212
answered Nov 25 '18 at 8:23
MartundMartund
1,433212
answered Nov 25 '18 at 8:23
MartundMartund
1,433212
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
|
show 5 more comments
$begingroup$
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
|
show 5 more comments
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StackExchange.helpers.onClickDraftSave('#login-link');
});
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StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
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StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
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Post as a guest
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Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
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