Calculate the area of all those squares, whose 2 corners are on the parabola and 2 remaining are on the x...
$begingroup$
You have a parabola $y=x^2-12x+36$. Calculate the area of all those squares, whose $2$ corners are on the given parabola and $2$ remaining are on the $x$-axis
real-analysis calculus analysis
edited Dec 13 '18 at 17:13
Sauhard Sharma
953318
asked Dec 13 '18 at 14:56
Bārbala Līna OstrovskaBārbala Līna Ostrovska
1
$endgroup$
add a comment |
$begingroup$
You have a parabola $y=x^2-12x+36$. Calculate the area of all those squares, whose $2$ corners are on the given parabola and $2$ remaining are on the $x$-axis
real-analysis calculus analysis
edited Dec 13 '18 at 17:13
Sauhard Sharma
953318
asked Dec 13 '18 at 14:56
Bārbala Līna OstrovskaBārbala Līna Ostrovska
1
$endgroup$
$begingroup$
Welcome to MSE. Please use MathJax to write questions. Also it is appreciated if you show what approach you have taken and where you got stuck
$endgroup$
– Sauhard Sharma
Dec 13 '18 at 15:08
add a comment |
$begingroup$
You have a parabola $y=x^2-12x+36$. Calculate the area of all those squares, whose $2$ corners are on the given parabola and $2$ remaining are on the $x$-axis
real-analysis calculus analysis
edited Dec 13 '18 at 17:13
Sauhard Sharma
953318
asked Dec 13 '18 at 14:56
Bārbala Līna OstrovskaBārbala Līna Ostrovska
1
$endgroup$
You have a parabola $y=x^2-12x+36$. Calculate the area of all those squares, whose $2$ corners are on the given parabola and $2$ remaining are on the $x$-axis
real-analysis calculus analysis
real-analysis calculus analysis
edited Dec 13 '18 at 17:13
Sauhard Sharma
953318
asked Dec 13 '18 at 14:56
Bārbala Līna OstrovskaBārbala Līna Ostrovska
1
edited Dec 13 '18 at 17:13
Sauhard Sharma
953318
asked Dec 13 '18 at 14:56
Bārbala Līna OstrovskaBārbala Līna Ostrovska
1
edited Dec 13 '18 at 17:13
Sauhard Sharma
953318
edited Dec 13 '18 at 17:13
Sauhard Sharma
953318
edited Dec 13 '18 at 17:13
Sauhard Sharma
953318
953318
asked Dec 13 '18 at 14:56
Bārbala Līna OstrovskaBārbala Līna Ostrovska
1
asked Dec 13 '18 at 14:56
Bārbala Līna OstrovskaBārbala Līna Ostrovska
1
asked Dec 13 '18 at 14:56
Bārbala Līna OstrovskaBārbala Līna Ostrovska
1
1
$begingroup$
Welcome to MSE. Please use MathJax to write questions. Also it is appreciated if you show what approach you have taken and where you got stuck
$endgroup$
– Sauhard Sharma
Dec 13 '18 at 15:08
add a comment |
$begingroup$
Welcome to MSE. Please use MathJax to write questions. Also it is appreciated if you show what approach you have taken and where you got stuck
$endgroup$
– Sauhard Sharma
Dec 13 '18 at 15:08
$begingroup$
Welcome to MSE. Please use MathJax to write questions. Also it is appreciated if you show what approach you have taken and where you got stuck
$endgroup$
– Sauhard Sharma
Dec 13 '18 at 15:08
$begingroup$
Welcome to MSE. Please use MathJax to write questions. Also it is appreciated if you show what approach you have taken and where you got stuck
$endgroup$
– Sauhard Sharma
Dec 13 '18 at 15:08
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You can start with the fact that if there is a square it has to be centred around the parabola. This means that it's around the point $(6,0)$. So we assume one corner is some $h$ distance to the left of it and one is $h$ distance to the right. This gives us the points $(6- h,0)$ and $(6+h,0)$. Now for it to be a square two vertical lines will go up from this point to meet the parabola. The vertical distance must be equal to the total horizontal distance which is $(6+h)-(6-h)=2h$.
To find out the vertical distance evaluate where the vertical lines will meet the parabola by putting in the $x=6-h$ and $x=6+h$.
$$y= (x-6)^2=(6-h-6)^2=h^2$$
So the vertical distance is $h^2$. Equating the vertical and the horizontal distances, we get
$$2h=h^2$$
$$h=2$$
and
$$h=0$$
$h=0$ is not allowed as it gives us a single point for all corners. When we put $h=2$, we get the $x$-axis points as $(4,0)$ and $(8,0)$. So the side of a square is $d=8-4=4$. Hence the area is $A=4cdot4=16$
answered Dec 13 '18 at 17:04
Sauhard SharmaSauhard Sharma
953318
$endgroup$
add a comment |
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
});
}
});
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%2f3038115%2fcalculate-the-area-of-all-those-squares-whose-2-corners-are-on-the-parabola-and%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 can start with the fact that if there is a square it has to be centred around the parabola. This means that it's around the point $(6,0)$. So we assume one corner is some $h$ distance to the left of it and one is $h$ distance to the right. This gives us the points $(6- h,0)$ and $(6+h,0)$. Now for it to be a square two vertical lines will go up from this point to meet the parabola. The vertical distance must be equal to the total horizontal distance which is $(6+h)-(6-h)=2h$.
To find out the vertical distance evaluate where the vertical lines will meet the parabola by putting in the $x=6-h$ and $x=6+h$.
$$y= (x-6)^2=(6-h-6)^2=h^2$$
So the vertical distance is $h^2$. Equating the vertical and the horizontal distances, we get
$$2h=h^2$$
$$h=2$$
and
$$h=0$$
$h=0$ is not allowed as it gives us a single point for all corners. When we put $h=2$, we get the $x$-axis points as $(4,0)$ and $(8,0)$. So the side of a square is $d=8-4=4$. Hence the area is $A=4cdot4=16$
answered Dec 13 '18 at 17:04
Sauhard SharmaSauhard Sharma
953318
$endgroup$
add a comment |
$begingroup$
You can start with the fact that if there is a square it has to be centred around the parabola. This means that it's around the point $(6,0)$. So we assume one corner is some $h$ distance to the left of it and one is $h$ distance to the right. This gives us the points $(6- h,0)$ and $(6+h,0)$. Now for it to be a square two vertical lines will go up from this point to meet the parabola. The vertical distance must be equal to the total horizontal distance which is $(6+h)-(6-h)=2h$.
To find out the vertical distance evaluate where the vertical lines will meet the parabola by putting in the $x=6-h$ and $x=6+h$.
$$y= (x-6)^2=(6-h-6)^2=h^2$$
So the vertical distance is $h^2$. Equating the vertical and the horizontal distances, we get
$$2h=h^2$$
$$h=2$$
and
$$h=0$$
$h=0$ is not allowed as it gives us a single point for all corners. When we put $h=2$, we get the $x$-axis points as $(4,0)$ and $(8,0)$. So the side of a square is $d=8-4=4$. Hence the area is $A=4cdot4=16$
answered Dec 13 '18 at 17:04
Sauhard SharmaSauhard Sharma
953318
$endgroup$
add a comment |
$begingroup$
You can start with the fact that if there is a square it has to be centred around the parabola. This means that it's around the point $(6,0)$. So we assume one corner is some $h$ distance to the left of it and one is $h$ distance to the right. This gives us the points $(6- h,0)$ and $(6+h,0)$. Now for it to be a square two vertical lines will go up from this point to meet the parabola. The vertical distance must be equal to the total horizontal distance which is $(6+h)-(6-h)=2h$.
To find out the vertical distance evaluate where the vertical lines will meet the parabola by putting in the $x=6-h$ and $x=6+h$.
$$y= (x-6)^2=(6-h-6)^2=h^2$$
So the vertical distance is $h^2$. Equating the vertical and the horizontal distances, we get
$$2h=h^2$$
$$h=2$$
and
$$h=0$$
$h=0$ is not allowed as it gives us a single point for all corners. When we put $h=2$, we get the $x$-axis points as $(4,0)$ and $(8,0)$. So the side of a square is $d=8-4=4$. Hence the area is $A=4cdot4=16$
answered Dec 13 '18 at 17:04
Sauhard SharmaSauhard Sharma
953318
$endgroup$
You can start with the fact that if there is a square it has to be centred around the parabola. This means that it's around the point $(6,0)$. So we assume one corner is some $h$ distance to the left of it and one is $h$ distance to the right. This gives us the points $(6- h,0)$ and $(6+h,0)$. Now for it to be a square two vertical lines will go up from this point to meet the parabola. The vertical distance must be equal to the total horizontal distance which is $(6+h)-(6-h)=2h$.
To find out the vertical distance evaluate where the vertical lines will meet the parabola by putting in the $x=6-h$ and $x=6+h$.
$$y= (x-6)^2=(6-h-6)^2=h^2$$
So the vertical distance is $h^2$. Equating the vertical and the horizontal distances, we get
$$2h=h^2$$
$$h=2$$
and
$$h=0$$
$h=0$ is not allowed as it gives us a single point for all corners. When we put $h=2$, we get the $x$-axis points as $(4,0)$ and $(8,0)$. So the side of a square is $d=8-4=4$. Hence the area is $A=4cdot4=16$
answered Dec 13 '18 at 17:04
Sauhard SharmaSauhard Sharma
953318
answered Dec 13 '18 at 17:04
Sauhard SharmaSauhard Sharma
953318
answered Dec 13 '18 at 17:04
Sauhard SharmaSauhard Sharma
953318
answered Dec 13 '18 at 17:04
Sauhard SharmaSauhard Sharma
953318
953318
add a comment |
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%2f3038115%2fcalculate-the-area-of-all-those-squares-whose-2-corners-are-on-the-parabola-and%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
$begingroup$
Welcome to MSE. Please use MathJax to write questions. Also it is appreciated if you show what approach you have taken and where you got stuck
$endgroup$
– Sauhard Sharma
Dec 13 '18 at 15:08