How to prove the triangle is isosceles and determine its area
$begingroup$
Let $P(a,b)$ be a point on the curve $y = frac{1}{x}$ in first quadrant and let the tangent line at $P$ intersect the $x$-axis at a point $A$. Show that triangle $AOP$ is isosceles and determine its area, where $O$ is the origin.
How do I start to attempt this?
analytic-geometry plane-geometry
edited Dec 14 '18 at 14:05
amWhy
1
asked Dec 13 '18 at 14:14
Valèt QuasarValèt Quasar
32
$endgroup$
add a comment |
$begingroup$
Let $P(a,b)$ be a point on the curve $y = frac{1}{x}$ in first quadrant and let the tangent line at $P$ intersect the $x$-axis at a point $A$. Show that triangle $AOP$ is isosceles and determine its area, where $O$ is the origin.
How do I start to attempt this?
analytic-geometry plane-geometry
edited Dec 14 '18 at 14:05
amWhy
1
asked Dec 13 '18 at 14:14
Valèt QuasarValèt Quasar
32
$endgroup$
1
$begingroup$
Have you tried to find the tangent line in terms of $P$ yet?
$endgroup$
– SmileyCraft
Dec 13 '18 at 14:18
$begingroup$
Very good. So the slope of $PA$ is $-1/a^2$. Now if you also calculate the slope of $OP$ you should notice something which makes the triangle very symmetric.
$endgroup$
– SmileyCraft
Dec 13 '18 at 15:04
$begingroup$
So essentially, the slope of $OP$ is just the inverse of $PA$? @SmileyCraft
$endgroup$
– Valèt Quasar
Dec 13 '18 at 15:10
$begingroup$
Right, the additive inverse that is. Because $O$ and $A$ lie on the $x$-axis, it follows that $|PA|$=$|OP|$.
$endgroup$
– SmileyCraft
Dec 13 '18 at 15:12
$begingroup$
I see now. Thank you so much for your assistance! @SmileyCraft
$endgroup$
– Valèt Quasar
Dec 13 '18 at 15:13
add a comment |
$begingroup$
Let $P(a,b)$ be a point on the curve $y = frac{1}{x}$ in first quadrant and let the tangent line at $P$ intersect the $x$-axis at a point $A$. Show that triangle $AOP$ is isosceles and determine its area, where $O$ is the origin.
How do I start to attempt this?
analytic-geometry plane-geometry
edited Dec 14 '18 at 14:05
amWhy
1
asked Dec 13 '18 at 14:14
Valèt QuasarValèt Quasar
32
$endgroup$
Let $P(a,b)$ be a point on the curve $y = frac{1}{x}$ in first quadrant and let the tangent line at $P$ intersect the $x$-axis at a point $A$. Show that triangle $AOP$ is isosceles and determine its area, where $O$ is the origin.
How do I start to attempt this?
analytic-geometry plane-geometry
analytic-geometry plane-geometry
edited Dec 14 '18 at 14:05
amWhy
1
asked Dec 13 '18 at 14:14
Valèt QuasarValèt Quasar
32
edited Dec 14 '18 at 14:05
amWhy
1
asked Dec 13 '18 at 14:14
Valèt QuasarValèt Quasar
32
edited Dec 14 '18 at 14:05
amWhy
1
edited Dec 14 '18 at 14:05
amWhy
1
edited Dec 14 '18 at 14:05
amWhy
1
1
asked Dec 13 '18 at 14:14
Valèt QuasarValèt Quasar
32
asked Dec 13 '18 at 14:14
Valèt QuasarValèt Quasar
32
asked Dec 13 '18 at 14:14
Valèt QuasarValèt Quasar
32
32
1
$begingroup$
Have you tried to find the tangent line in terms of $P$ yet?
$endgroup$
– SmileyCraft
Dec 13 '18 at 14:18
$begingroup$
Very good. So the slope of $PA$ is $-1/a^2$. Now if you also calculate the slope of $OP$ you should notice something which makes the triangle very symmetric.
$endgroup$
– SmileyCraft
Dec 13 '18 at 15:04
$begingroup$
So essentially, the slope of $OP$ is just the inverse of $PA$? @SmileyCraft
$endgroup$
– Valèt Quasar
Dec 13 '18 at 15:10
$begingroup$
Right, the additive inverse that is. Because $O$ and $A$ lie on the $x$-axis, it follows that $|PA|$=$|OP|$.
$endgroup$
– SmileyCraft
Dec 13 '18 at 15:12
$begingroup$
I see now. Thank you so much for your assistance! @SmileyCraft
$endgroup$
– Valèt Quasar
Dec 13 '18 at 15:13
add a comment |
1
$begingroup$
Have you tried to find the tangent line in terms of $P$ yet?
$endgroup$
– SmileyCraft
Dec 13 '18 at 14:18
$begingroup$
Very good. So the slope of $PA$ is $-1/a^2$. Now if you also calculate the slope of $OP$ you should notice something which makes the triangle very symmetric.
$endgroup$
– SmileyCraft
Dec 13 '18 at 15:04
$begingroup$
So essentially, the slope of $OP$ is just the inverse of $PA$? @SmileyCraft
$endgroup$
– Valèt Quasar
Dec 13 '18 at 15:10
$begingroup$
Right, the additive inverse that is. Because $O$ and $A$ lie on the $x$-axis, it follows that $|PA|$=$|OP|$.
$endgroup$
– SmileyCraft
Dec 13 '18 at 15:12
$begingroup$
I see now. Thank you so much for your assistance! @SmileyCraft
$endgroup$
– Valèt Quasar
Dec 13 '18 at 15:13
1
1
$begingroup$
Have you tried to find the tangent line in terms of $P$ yet?
$endgroup$
– SmileyCraft
Dec 13 '18 at 14:18
$begingroup$
Have you tried to find the tangent line in terms of $P$ yet?
$endgroup$
– SmileyCraft
Dec 13 '18 at 14:18
$begingroup$
Very good. So the slope of $PA$ is $-1/a^2$. Now if you also calculate the slope of $OP$ you should notice something which makes the triangle very symmetric.
$endgroup$
– SmileyCraft
Dec 13 '18 at 15:04
$begingroup$
Very good. So the slope of $PA$ is $-1/a^2$. Now if you also calculate the slope of $OP$ you should notice something which makes the triangle very symmetric.
$endgroup$
– SmileyCraft
Dec 13 '18 at 15:04
$begingroup$
So essentially, the slope of $OP$ is just the inverse of $PA$? @SmileyCraft
$endgroup$
– Valèt Quasar
Dec 13 '18 at 15:10
$begingroup$
So essentially, the slope of $OP$ is just the inverse of $PA$? @SmileyCraft
$endgroup$
– Valèt Quasar
Dec 13 '18 at 15:10
$begingroup$
Right, the additive inverse that is. Because $O$ and $A$ lie on the $x$-axis, it follows that $|PA|$=$|OP|$.
$endgroup$
– SmileyCraft
Dec 13 '18 at 15:12
$begingroup$
Right, the additive inverse that is. Because $O$ and $A$ lie on the $x$-axis, it follows that $|PA|$=$|OP|$.
$endgroup$
– SmileyCraft
Dec 13 '18 at 15:12
$begingroup$
I see now. Thank you so much for your assistance! @SmileyCraft
$endgroup$
– Valèt Quasar
Dec 13 '18 at 15:13
$begingroup$
I see now. Thank you so much for your assistance! @SmileyCraft
$endgroup$
– Valèt Quasar
Dec 13 '18 at 15:13
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
To start approaching the problem, first notice that if P lies on the curve then $P (a, b) = P (a, frac{1}{a})$.
Secondly, find the slope of the tangent line at $P$, which is precisely $-frac{1}{a^2}$
Thus equation of the tangent line through $P$ is given by
$$(y - frac{1}{a}) = (-frac{1}{a^2})(x-a) implies a^2y = 2a - x$$
The above line intersects the $x$-axis only at $(2a, 0)$, thus $ A = (2a, 0)$.
Therefore the coordinates of $Delta AOP$ are precisely: $$A (2a, 0), O(0, 0), P (a, frac{1}{a})$$
making $AOP$ isosceles. Moreover, the area of $Delta AOP = frac{1}{2} cdot 2a cdot frac{1}{a} = 1$ unit.
answered Dec 13 '18 at 20:45
thesagniksahathesagniksaha
164115
$endgroup$
add a comment |
$begingroup$
Hint:
WLOG $b=1/a>0$
$$xy-1implies xdfrac{dy}{dx}+y=0impliesdfrac{dy}{dx}_{text{ at }(a,b)}=-dfrac{1/a}a=-dfrac1{a^2}$$
The equation of the tangent at $P,$ $$dfrac{y-1/a}{x-a}=-dfrac1{a^2}iff x+a^2y=2a$$
As $x$ axis, $y=0,$ the intersection will be $A(2a,0)$
$$|AP|=?, |OP|=?$$
answered Dec 13 '18 at 14:22
lab bhattacharjeelab bhattacharjee
225k15157275
$endgroup$
add a comment |
$begingroup$
The tangent line to $y=frac 1x$ at $P(x_0,y_0)$ is:
$$y=y_0+y'(x_0)(x-x_0)=frac 1{x_0}-frac 1{x_0^2}(x-x_0).$$
The point $A(x_1,0)$ has coordinate $x_1$:
$$0=frac 1{x_0}-frac 1{x_0^2}(x_1-x_0) Rightarrow x_1=2x_0,$$
which shows the triangle $PAO$ is isosceles, because the altitude $PB$ is the median too.
Now the area of $Delta PAO$ is:
$$S=frac 12 cdot AOcdot PB=frac12 cdot 2x_0 cdot frac1{x_0}=1.$$
answered Dec 13 '18 at 16:24
farruhotafarruhota
20.2k2738
$endgroup$
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%2f3038053%2fhow-to-prove-the-triangle-is-isosceles-and-determine-its-area%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
To start approaching the problem, first notice that if P lies on the curve then $P (a, b) = P (a, frac{1}{a})$.
Secondly, find the slope of the tangent line at $P$, which is precisely $-frac{1}{a^2}$
Thus equation of the tangent line through $P$ is given by
$$(y - frac{1}{a}) = (-frac{1}{a^2})(x-a) implies a^2y = 2a - x$$
The above line intersects the $x$-axis only at $(2a, 0)$, thus $ A = (2a, 0)$.
Therefore the coordinates of $Delta AOP$ are precisely: $$A (2a, 0), O(0, 0), P (a, frac{1}{a})$$
making $AOP$ isosceles. Moreover, the area of $Delta AOP = frac{1}{2} cdot 2a cdot frac{1}{a} = 1$ unit.
answered Dec 13 '18 at 20:45
thesagniksahathesagniksaha
164115
$endgroup$
add a comment |
$begingroup$
To start approaching the problem, first notice that if P lies on the curve then $P (a, b) = P (a, frac{1}{a})$.
Secondly, find the slope of the tangent line at $P$, which is precisely $-frac{1}{a^2}$
Thus equation of the tangent line through $P$ is given by
$$(y - frac{1}{a}) = (-frac{1}{a^2})(x-a) implies a^2y = 2a - x$$
The above line intersects the $x$-axis only at $(2a, 0)$, thus $ A = (2a, 0)$.
Therefore the coordinates of $Delta AOP$ are precisely: $$A (2a, 0), O(0, 0), P (a, frac{1}{a})$$
making $AOP$ isosceles. Moreover, the area of $Delta AOP = frac{1}{2} cdot 2a cdot frac{1}{a} = 1$ unit.
answered Dec 13 '18 at 20:45
thesagniksahathesagniksaha
164115
$endgroup$
add a comment |
$begingroup$
To start approaching the problem, first notice that if P lies on the curve then $P (a, b) = P (a, frac{1}{a})$.
Secondly, find the slope of the tangent line at $P$, which is precisely $-frac{1}{a^2}$
Thus equation of the tangent line through $P$ is given by
$$(y - frac{1}{a}) = (-frac{1}{a^2})(x-a) implies a^2y = 2a - x$$
The above line intersects the $x$-axis only at $(2a, 0)$, thus $ A = (2a, 0)$.
Therefore the coordinates of $Delta AOP$ are precisely: $$A (2a, 0), O(0, 0), P (a, frac{1}{a})$$
making $AOP$ isosceles. Moreover, the area of $Delta AOP = frac{1}{2} cdot 2a cdot frac{1}{a} = 1$ unit.
answered Dec 13 '18 at 20:45
thesagniksahathesagniksaha
164115
$endgroup$
To start approaching the problem, first notice that if P lies on the curve then $P (a, b) = P (a, frac{1}{a})$.
Secondly, find the slope of the tangent line at $P$, which is precisely $-frac{1}{a^2}$
Thus equation of the tangent line through $P$ is given by
$$(y - frac{1}{a}) = (-frac{1}{a^2})(x-a) implies a^2y = 2a - x$$
The above line intersects the $x$-axis only at $(2a, 0)$, thus $ A = (2a, 0)$.
Therefore the coordinates of $Delta AOP$ are precisely: $$A (2a, 0), O(0, 0), P (a, frac{1}{a})$$
making $AOP$ isosceles. Moreover, the area of $Delta AOP = frac{1}{2} cdot 2a cdot frac{1}{a} = 1$ unit.
answered Dec 13 '18 at 20:45
thesagniksahathesagniksaha
164115
edited Dec 14 '18 at 13:54
answered Dec 13 '18 at 20:45
thesagniksahathesagniksaha
164115
answered Dec 13 '18 at 20:45
thesagniksahathesagniksaha
164115
answered Dec 13 '18 at 20:45
thesagniksahathesagniksaha
164115
164115
add a comment |
add a comment |
$begingroup$
Hint:
WLOG $b=1/a>0$
$$xy-1implies xdfrac{dy}{dx}+y=0impliesdfrac{dy}{dx}_{text{ at }(a,b)}=-dfrac{1/a}a=-dfrac1{a^2}$$
The equation of the tangent at $P,$ $$dfrac{y-1/a}{x-a}=-dfrac1{a^2}iff x+a^2y=2a$$
As $x$ axis, $y=0,$ the intersection will be $A(2a,0)$
$$|AP|=?, |OP|=?$$
answered Dec 13 '18 at 14:22
lab bhattacharjeelab bhattacharjee
225k15157275
$endgroup$
add a comment |
$begingroup$
Hint:
WLOG $b=1/a>0$
$$xy-1implies xdfrac{dy}{dx}+y=0impliesdfrac{dy}{dx}_{text{ at }(a,b)}=-dfrac{1/a}a=-dfrac1{a^2}$$
The equation of the tangent at $P,$ $$dfrac{y-1/a}{x-a}=-dfrac1{a^2}iff x+a^2y=2a$$
As $x$ axis, $y=0,$ the intersection will be $A(2a,0)$
$$|AP|=?, |OP|=?$$
answered Dec 13 '18 at 14:22
lab bhattacharjeelab bhattacharjee
225k15157275
$endgroup$
add a comment |
$begingroup$
Hint:
WLOG $b=1/a>0$
$$xy-1implies xdfrac{dy}{dx}+y=0impliesdfrac{dy}{dx}_{text{ at }(a,b)}=-dfrac{1/a}a=-dfrac1{a^2}$$
The equation of the tangent at $P,$ $$dfrac{y-1/a}{x-a}=-dfrac1{a^2}iff x+a^2y=2a$$
As $x$ axis, $y=0,$ the intersection will be $A(2a,0)$
$$|AP|=?, |OP|=?$$
answered Dec 13 '18 at 14:22
lab bhattacharjeelab bhattacharjee
225k15157275
$endgroup$
Hint:
WLOG $b=1/a>0$
$$xy-1implies xdfrac{dy}{dx}+y=0impliesdfrac{dy}{dx}_{text{ at }(a,b)}=-dfrac{1/a}a=-dfrac1{a^2}$$
The equation of the tangent at $P,$ $$dfrac{y-1/a}{x-a}=-dfrac1{a^2}iff x+a^2y=2a$$
As $x$ axis, $y=0,$ the intersection will be $A(2a,0)$
$$|AP|=?, |OP|=?$$
answered Dec 13 '18 at 14:22
lab bhattacharjeelab bhattacharjee
225k15157275
answered Dec 13 '18 at 14:22
lab bhattacharjeelab bhattacharjee
225k15157275
answered Dec 13 '18 at 14:22
lab bhattacharjeelab bhattacharjee
225k15157275
answered Dec 13 '18 at 14:22
lab bhattacharjeelab bhattacharjee
225k15157275
225k15157275
add a comment |
add a comment |
$begingroup$
The tangent line to $y=frac 1x$ at $P(x_0,y_0)$ is:
$$y=y_0+y'(x_0)(x-x_0)=frac 1{x_0}-frac 1{x_0^2}(x-x_0).$$
The point $A(x_1,0)$ has coordinate $x_1$:
$$0=frac 1{x_0}-frac 1{x_0^2}(x_1-x_0) Rightarrow x_1=2x_0,$$
which shows the triangle $PAO$ is isosceles, because the altitude $PB$ is the median too.
Now the area of $Delta PAO$ is:
$$S=frac 12 cdot AOcdot PB=frac12 cdot 2x_0 cdot frac1{x_0}=1.$$
answered Dec 13 '18 at 16:24
farruhotafarruhota
20.2k2738
$endgroup$
add a comment |
$begingroup$
The tangent line to $y=frac 1x$ at $P(x_0,y_0)$ is:
$$y=y_0+y'(x_0)(x-x_0)=frac 1{x_0}-frac 1{x_0^2}(x-x_0).$$
The point $A(x_1,0)$ has coordinate $x_1$:
$$0=frac 1{x_0}-frac 1{x_0^2}(x_1-x_0) Rightarrow x_1=2x_0,$$
which shows the triangle $PAO$ is isosceles, because the altitude $PB$ is the median too.
Now the area of $Delta PAO$ is:
$$S=frac 12 cdot AOcdot PB=frac12 cdot 2x_0 cdot frac1{x_0}=1.$$
answered Dec 13 '18 at 16:24
farruhotafarruhota
20.2k2738
$endgroup$
add a comment |
$begingroup$
The tangent line to $y=frac 1x$ at $P(x_0,y_0)$ is:
$$y=y_0+y'(x_0)(x-x_0)=frac 1{x_0}-frac 1{x_0^2}(x-x_0).$$
The point $A(x_1,0)$ has coordinate $x_1$:
$$0=frac 1{x_0}-frac 1{x_0^2}(x_1-x_0) Rightarrow x_1=2x_0,$$
which shows the triangle $PAO$ is isosceles, because the altitude $PB$ is the median too.
Now the area of $Delta PAO$ is:
$$S=frac 12 cdot AOcdot PB=frac12 cdot 2x_0 cdot frac1{x_0}=1.$$
answered Dec 13 '18 at 16:24
farruhotafarruhota
20.2k2738
$endgroup$
The tangent line to $y=frac 1x$ at $P(x_0,y_0)$ is:
$$y=y_0+y'(x_0)(x-x_0)=frac 1{x_0}-frac 1{x_0^2}(x-x_0).$$
The point $A(x_1,0)$ has coordinate $x_1$:
$$0=frac 1{x_0}-frac 1{x_0^2}(x_1-x_0) Rightarrow x_1=2x_0,$$
which shows the triangle $PAO$ is isosceles, because the altitude $PB$ is the median too.
Now the area of $Delta PAO$ is:
$$S=frac 12 cdot AOcdot PB=frac12 cdot 2x_0 cdot frac1{x_0}=1.$$
answered Dec 13 '18 at 16:24
farruhotafarruhota
20.2k2738
answered Dec 13 '18 at 16:24
farruhotafarruhota
20.2k2738
answered Dec 13 '18 at 16:24
farruhotafarruhota
20.2k2738
answered Dec 13 '18 at 16:24
farruhotafarruhota
20.2k2738
20.2k2738
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%2f3038053%2fhow-to-prove-the-triangle-is-isosceles-and-determine-its-area%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
1
$begingroup$
Have you tried to find the tangent line in terms of $P$ yet?
$endgroup$
– SmileyCraft
Dec 13 '18 at 14:18
$begingroup$
Very good. So the slope of $PA$ is $-1/a^2$. Now if you also calculate the slope of $OP$ you should notice something which makes the triangle very symmetric.
$endgroup$
– SmileyCraft
Dec 13 '18 at 15:04
$begingroup$
So essentially, the slope of $OP$ is just the inverse of $PA$? @SmileyCraft
$endgroup$
– Valèt Quasar
Dec 13 '18 at 15:10
$begingroup$
Right, the additive inverse that is. Because $O$ and $A$ lie on the $x$-axis, it follows that $|PA|$=$|OP|$.
$endgroup$
– SmileyCraft
Dec 13 '18 at 15:12
$begingroup$
I see now. Thank you so much for your assistance! @SmileyCraft
$endgroup$
– Valèt Quasar
Dec 13 '18 at 15:13