Mixing
Conic sections: Parabola - What is $p$?
$begingroup$
Help my teacher says $p$ can't be negative because it's distance.
I watched TOCT's tutorial (The Organic Chemistry Tutor) in YouTube about parabola and he said "$(x-h)^2 = 4p(y-k)$ if $p$ is positive the parabola opens upward, if negative parabola opens downward. $(y-k)^2 = 4p(x-h) $ if $p$ is positive the parabola opens to the right, if $p$ negative parabola opens to the left."
So I tried answering the question my teacher gave me:
"Find the equation of the parabola with Vertex$(1,2)$ and Focus$(1, -8)$"
I tried graphing the given points and found out the focus is below the vertex.
I used distance formula to find out $p$: $p=pm 10$
Based on what I watched in YouTube if the parabola opens downwards, $p$ is negative:
$p=-10$
But then my teacher said $p$ can't be negative so I get the absolute value so:
$p=10$
I tried to get the equation of the parabola using $(x-h)^2 = 4p(y-k)$:
With $p$ positive
$(x-1)^2 = 4(10)(y-2)$
$(x-1)^2 = 40(y-2)$
With $p$ negative
$(x-1)^2 = 4(-10)(y-2)$
$(x-1)^2 = -40(y-2)$
I tried to find the directrix with the formula $y=k-p$:
With $p$ positive
$y=2-(+10)$
$y=2-10$
$y=-8$
With $p$ negative
$y=2-(-10)$
$y=2+10$
$y=12$
When I graphed the directrix I think the negative one makes more sense so I am soooo confused right now
I wanna know what is the correct formula. Which equation is the correct one? Which directrix is correct? What is $p$? I also want to know how to get the ELR (end of latus rectum which is $2p$ units away from the focus) Help me understand all of this better!!
conic-sections
edited Jan 29 at 14:41
Henrik
6,04592030
asked Jan 29 at 12:59
user639467user639467
82
$endgroup$
add a comment |
$begingroup$
Help my teacher says $p$ can't be negative because it's distance.
I watched TOCT's tutorial (The Organic Chemistry Tutor) in YouTube about parabola and he said "$(x-h)^2 = 4p(y-k)$ if $p$ is positive the parabola opens upward, if negative parabola opens downward. $(y-k)^2 = 4p(x-h) $ if $p$ is positive the parabola opens to the right, if $p$ negative parabola opens to the left."
So I tried answering the question my teacher gave me:
"Find the equation of the parabola with Vertex$(1,2)$ and Focus$(1, -8)$"
I tried graphing the given points and found out the focus is below the vertex.
I used distance formula to find out $p$: $p=pm 10$
Based on what I watched in YouTube if the parabola opens downwards, $p$ is negative:
$p=-10$
But then my teacher said $p$ can't be negative so I get the absolute value so:
$p=10$
I tried to get the equation of the parabola using $(x-h)^2 = 4p(y-k)$:
With $p$ positive
$(x-1)^2 = 4(10)(y-2)$
$(x-1)^2 = 40(y-2)$
With $p$ negative
$(x-1)^2 = 4(-10)(y-2)$
$(x-1)^2 = -40(y-2)$
I tried to find the directrix with the formula $y=k-p$:
With $p$ positive
$y=2-(+10)$
$y=2-10$
$y=-8$
With $p$ negative
$y=2-(-10)$
$y=2+10$
$y=12$
When I graphed the directrix I think the negative one makes more sense so I am soooo confused right now
I wanna know what is the correct formula. Which equation is the correct one? Which directrix is correct? What is $p$? I also want to know how to get the ELR (end of latus rectum which is $2p$ units away from the focus) Help me understand all of this better!!
conic-sections
edited Jan 29 at 14:41
Henrik
6,04592030
asked Jan 29 at 12:59
user639467user639467
82
$endgroup$
add a comment |
$begingroup$
Help my teacher says $p$ can't be negative because it's distance.
I watched TOCT's tutorial (The Organic Chemistry Tutor) in YouTube about parabola and he said "$(x-h)^2 = 4p(y-k)$ if $p$ is positive the parabola opens upward, if negative parabola opens downward. $(y-k)^2 = 4p(x-h) $ if $p$ is positive the parabola opens to the right, if $p$ negative parabola opens to the left."
So I tried answering the question my teacher gave me:
"Find the equation of the parabola with Vertex$(1,2)$ and Focus$(1, -8)$"
I tried graphing the given points and found out the focus is below the vertex.
I used distance formula to find out $p$: $p=pm 10$
Based on what I watched in YouTube if the parabola opens downwards, $p$ is negative:
$p=-10$
But then my teacher said $p$ can't be negative so I get the absolute value so:
$p=10$
I tried to get the equation of the parabola using $(x-h)^2 = 4p(y-k)$:
With $p$ positive
$(x-1)^2 = 4(10)(y-2)$
$(x-1)^2 = 40(y-2)$
With $p$ negative
$(x-1)^2 = 4(-10)(y-2)$
$(x-1)^2 = -40(y-2)$
I tried to find the directrix with the formula $y=k-p$:
With $p$ positive
$y=2-(+10)$
$y=2-10$
$y=-8$
With $p$ negative
$y=2-(-10)$
$y=2+10$
$y=12$
When I graphed the directrix I think the negative one makes more sense so I am soooo confused right now
I wanna know what is the correct formula. Which equation is the correct one? Which directrix is correct? What is $p$? I also want to know how to get the ELR (end of latus rectum which is $2p$ units away from the focus) Help me understand all of this better!!
conic-sections
edited Jan 29 at 14:41
Henrik
6,04592030
asked Jan 29 at 12:59
user639467user639467
82
$endgroup$
Help my teacher says $p$ can't be negative because it's distance.
I watched TOCT's tutorial (The Organic Chemistry Tutor) in YouTube about parabola and he said "$(x-h)^2 = 4p(y-k)$ if $p$ is positive the parabola opens upward, if negative parabola opens downward. $(y-k)^2 = 4p(x-h) $ if $p$ is positive the parabola opens to the right, if $p$ negative parabola opens to the left."
So I tried answering the question my teacher gave me:
"Find the equation of the parabola with Vertex$(1,2)$ and Focus$(1, -8)$"
I tried graphing the given points and found out the focus is below the vertex.
I used distance formula to find out $p$: $p=pm 10$
Based on what I watched in YouTube if the parabola opens downwards, $p$ is negative:
$p=-10$
But then my teacher said $p$ can't be negative so I get the absolute value so:
$p=10$
I tried to get the equation of the parabola using $(x-h)^2 = 4p(y-k)$:
With $p$ positive
$(x-1)^2 = 4(10)(y-2)$
$(x-1)^2 = 40(y-2)$
With $p$ negative
$(x-1)^2 = 4(-10)(y-2)$
$(x-1)^2 = -40(y-2)$
I tried to find the directrix with the formula $y=k-p$:
With $p$ positive
$y=2-(+10)$
$y=2-10$
$y=-8$
With $p$ negative
$y=2-(-10)$
$y=2+10$
$y=12$
When I graphed the directrix I think the negative one makes more sense so I am soooo confused right now
I wanna know what is the correct formula. Which equation is the correct one? Which directrix is correct? What is $p$? I also want to know how to get the ELR (end of latus rectum which is $2p$ units away from the focus) Help me understand all of this better!!
conic-sections
conic-sections
edited Jan 29 at 14:41
Henrik
6,04592030
asked Jan 29 at 12:59
user639467user639467
82
edited Jan 29 at 14:41
Henrik
6,04592030
asked Jan 29 at 12:59
user639467user639467
82
edited Jan 29 at 14:41
Henrik
6,04592030
edited Jan 29 at 14:41
Henrik
6,04592030
edited Jan 29 at 14:41
Henrik
6,04592030
6,04592030
asked Jan 29 at 12:59
user639467user639467
82
asked Jan 29 at 12:59
user639467user639467
82
asked Jan 29 at 12:59
user639467user639467
82
82
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You are right - since the vertex is above the focus, the parabola must open downwards. So $p=-10$ and the parabola has equation $(x-1)^2 = -40(y-2)$. Note that the right hand side is never negative because for all points on the parabola $y le 2$ (because the parabola opens downwards). The directrix is on the opposite side of the vertex to the focus, and is the line $y=12$.
The latus rectum is the horizontal chord of the parabola that passes through the focus, so it is part of the line with equation $y=-8$. This intersects the parabola at the two points where
$(x-1)^2 = -40(y-2) = -40(-8-2) = 400 \ Rightarrow x-1 = pm20 \ Rightarrow x = -19 text{ or } x=21$
answered Jan 29 at 14:19
gandalf61gandalf61
9,184825
$endgroup$
$begingroup$
This is really helpful! I also like what you did for finding the LR I didn't know that because the formula that I got from my teacher for finding LR was (h+2p, k+p) and (h-2p, k+p) and that is only for parabola facing up, there's like 6 more for facing down, right, and left. Are these formulas even necessary and is my teacher correct about having p as positive?
$endgroup$
– user639467
Jan 29 at 14:37
$begingroup$
For a downward opening parabola $p$ will be negative. The formula $(h pm 2p, k + p)$ for the ends of the latus rectum is stil correct for negative $p$ - in your example the ends of the latus rectum are at $(1 pm (-20), 2+(-10)) = (-19,-8) text{ or } (21,-8)$.
$endgroup$
– gandalf61
Jan 29 at 15:32
add a comment |
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
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3092141%2fconic-sections-parabola-what-is-p%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You are right - since the vertex is above the focus, the parabola must open downwards. So $p=-10$ and the parabola has equation $(x-1)^2 = -40(y-2)$. Note that the right hand side is never negative because for all points on the parabola $y le 2$ (because the parabola opens downwards). The directrix is on the opposite side of the vertex to the focus, and is the line $y=12$.
The latus rectum is the horizontal chord of the parabola that passes through the focus, so it is part of the line with equation $y=-8$. This intersects the parabola at the two points where
$(x-1)^2 = -40(y-2) = -40(-8-2) = 400 \ Rightarrow x-1 = pm20 \ Rightarrow x = -19 text{ or } x=21$
answered Jan 29 at 14:19
gandalf61gandalf61
9,184825
$endgroup$
$begingroup$
This is really helpful! I also like what you did for finding the LR I didn't know that because the formula that I got from my teacher for finding LR was (h+2p, k+p) and (h-2p, k+p) and that is only for parabola facing up, there's like 6 more for facing down, right, and left. Are these formulas even necessary and is my teacher correct about having p as positive?
$endgroup$
– user639467
Jan 29 at 14:37
$begingroup$
For a downward opening parabola $p$ will be negative. The formula $(h pm 2p, k + p)$ for the ends of the latus rectum is stil correct for negative $p$ - in your example the ends of the latus rectum are at $(1 pm (-20), 2+(-10)) = (-19,-8) text{ or } (21,-8)$.
$endgroup$
– gandalf61
Jan 29 at 15:32
add a comment |
$begingroup$
You are right - since the vertex is above the focus, the parabola must open downwards. So $p=-10$ and the parabola has equation $(x-1)^2 = -40(y-2)$. Note that the right hand side is never negative because for all points on the parabola $y le 2$ (because the parabola opens downwards). The directrix is on the opposite side of the vertex to the focus, and is the line $y=12$.
The latus rectum is the horizontal chord of the parabola that passes through the focus, so it is part of the line with equation $y=-8$. This intersects the parabola at the two points where
$(x-1)^2 = -40(y-2) = -40(-8-2) = 400 \ Rightarrow x-1 = pm20 \ Rightarrow x = -19 text{ or } x=21$
answered Jan 29 at 14:19
gandalf61gandalf61
9,184825
$endgroup$
$begingroup$
This is really helpful! I also like what you did for finding the LR I didn't know that because the formula that I got from my teacher for finding LR was (h+2p, k+p) and (h-2p, k+p) and that is only for parabola facing up, there's like 6 more for facing down, right, and left. Are these formulas even necessary and is my teacher correct about having p as positive?
$endgroup$
– user639467
Jan 29 at 14:37
$begingroup$
For a downward opening parabola $p$ will be negative. The formula $(h pm 2p, k + p)$ for the ends of the latus rectum is stil correct for negative $p$ - in your example the ends of the latus rectum are at $(1 pm (-20), 2+(-10)) = (-19,-8) text{ or } (21,-8)$.
$endgroup$
– gandalf61
Jan 29 at 15:32
add a comment |
$begingroup$
You are right - since the vertex is above the focus, the parabola must open downwards. So $p=-10$ and the parabola has equation $(x-1)^2 = -40(y-2)$. Note that the right hand side is never negative because for all points on the parabola $y le 2$ (because the parabola opens downwards). The directrix is on the opposite side of the vertex to the focus, and is the line $y=12$.
The latus rectum is the horizontal chord of the parabola that passes through the focus, so it is part of the line with equation $y=-8$. This intersects the parabola at the two points where
$(x-1)^2 = -40(y-2) = -40(-8-2) = 400 \ Rightarrow x-1 = pm20 \ Rightarrow x = -19 text{ or } x=21$
answered Jan 29 at 14:19
gandalf61gandalf61
9,184825
$endgroup$
You are right - since the vertex is above the focus, the parabola must open downwards. So $p=-10$ and the parabola has equation $(x-1)^2 = -40(y-2)$. Note that the right hand side is never negative because for all points on the parabola $y le 2$ (because the parabola opens downwards). The directrix is on the opposite side of the vertex to the focus, and is the line $y=12$.
The latus rectum is the horizontal chord of the parabola that passes through the focus, so it is part of the line with equation $y=-8$. This intersects the parabola at the two points where
$(x-1)^2 = -40(y-2) = -40(-8-2) = 400 \ Rightarrow x-1 = pm20 \ Rightarrow x = -19 text{ or } x=21$
answered Jan 29 at 14:19
gandalf61gandalf61
9,184825
answered Jan 29 at 14:19
gandalf61gandalf61
9,184825
answered Jan 29 at 14:19
gandalf61gandalf61
9,184825
answered Jan 29 at 14:19
gandalf61gandalf61
9,184825
9,184825
$begingroup$
This is really helpful! I also like what you did for finding the LR I didn't know that because the formula that I got from my teacher for finding LR was (h+2p, k+p) and (h-2p, k+p) and that is only for parabola facing up, there's like 6 more for facing down, right, and left. Are these formulas even necessary and is my teacher correct about having p as positive?
$endgroup$
– user639467
Jan 29 at 14:37
$begingroup$
For a downward opening parabola $p$ will be negative. The formula $(h pm 2p, k + p)$ for the ends of the latus rectum is stil correct for negative $p$ - in your example the ends of the latus rectum are at $(1 pm (-20), 2+(-10)) = (-19,-8) text{ or } (21,-8)$.
$endgroup$
– gandalf61
Jan 29 at 15:32
add a comment |
$begingroup$
This is really helpful! I also like what you did for finding the LR I didn't know that because the formula that I got from my teacher for finding LR was (h+2p, k+p) and (h-2p, k+p) and that is only for parabola facing up, there's like 6 more for facing down, right, and left. Are these formulas even necessary and is my teacher correct about having p as positive?
$endgroup$
– user639467
Jan 29 at 14:37
$begingroup$
For a downward opening parabola $p$ will be negative. The formula $(h pm 2p, k + p)$ for the ends of the latus rectum is stil correct for negative $p$ - in your example the ends of the latus rectum are at $(1 pm (-20), 2+(-10)) = (-19,-8) text{ or } (21,-8)$.
$endgroup$
– gandalf61
Jan 29 at 15:32
$begingroup$
This is really helpful! I also like what you did for finding the LR I didn't know that because the formula that I got from my teacher for finding LR was (h+2p, k+p) and (h-2p, k+p) and that is only for parabola facing up, there's like 6 more for facing down, right, and left. Are these formulas even necessary and is my teacher correct about having p as positive?
$endgroup$
– user639467
Jan 29 at 14:37
$begingroup$
This is really helpful! I also like what you did for finding the LR I didn't know that because the formula that I got from my teacher for finding LR was (h+2p, k+p) and (h-2p, k+p) and that is only for parabola facing up, there's like 6 more for facing down, right, and left. Are these formulas even necessary and is my teacher correct about having p as positive?
$endgroup$
– user639467
Jan 29 at 14:37
$begingroup$
For a downward opening parabola $p$ will be negative. The formula $(h pm 2p, k + p)$ for the ends of the latus rectum is stil correct for negative $p$ - in your example the ends of the latus rectum are at $(1 pm (-20), 2+(-10)) = (-19,-8) text{ or } (21,-8)$.
$endgroup$
– gandalf61
Jan 29 at 15:32
$begingroup$
For a downward opening parabola $p$ will be negative. The formula $(h pm 2p, k + p)$ for the ends of the latus rectum is stil correct for negative $p$ - in your example the ends of the latus rectum are at $(1 pm (-20), 2+(-10)) = (-19,-8) text{ or } (21,-8)$.
$endgroup$
– gandalf61
Jan 29 at 15:32
add a comment |
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.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3092141%2fconic-sections-parabola-what-is-p%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
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
