Mixing
The area of the triangle formed by the coordinate axes and tangent at vertex to the parabola whose focus is...
$begingroup$
The area of the triangle formed by the coordinate axes and tangent at vertex to the parabola whose focus is $(3,4)$ and tangents at $x=0$ and $y=0$ is?
I know how to do this, assume an equation for parabola with axis as $y=4/3 x$ and do all the procedure. But, this isn't how it is really to be solved. There's a sleek way which I cannot find.
conic-sections
asked Jan 3 at 12:13
Ice InkberryIce Inkberry
336112
$endgroup$
add a comment |
$begingroup$
The area of the triangle formed by the coordinate axes and tangent at vertex to the parabola whose focus is $(3,4)$ and tangents at $x=0$ and $y=0$ is?
I know how to do this, assume an equation for parabola with axis as $y=4/3 x$ and do all the procedure. But, this isn't how it is really to be solved. There's a sleek way which I cannot find.
conic-sections
asked Jan 3 at 12:13
Ice InkberryIce Inkberry
336112
$endgroup$
$begingroup$
This question is not well posed. There exist an infinite number of parabolas with focus at (3, 4). More information is needed. And "y= 4/3x" is NOT a parabola. A parabola with vertical line of symmetry can be written as $y= a(x- h)^2+ k$ and has focus $left(h, k+ frac{1}{4a}right)$. Saying that the focus is at (3, 4) means h= 3 and $k+ frac{1}{4a}= 4$ so $k= 4-frac{1}{4a}$. A parabola of the form $y= a(x- 3)^2+ 4- frac{1}{4a}$, for any a, has focus (3, 4).
$endgroup$
– user247327
Jan 3 at 12:19
$begingroup$
@user247327 No, the parabola has x=y=0 as tangents. Also, $y=4/3x$ is axis of the parabola.
$endgroup$
– Ice Inkberry
Jan 3 at 12:21
1
$begingroup$
How do you know that the parabola’s axis passes through the origin as you claim with that equation of the axis?
$endgroup$
– amd
Jan 3 at 21:46
add a comment |
$begingroup$
The area of the triangle formed by the coordinate axes and tangent at vertex to the parabola whose focus is $(3,4)$ and tangents at $x=0$ and $y=0$ is?
I know how to do this, assume an equation for parabola with axis as $y=4/3 x$ and do all the procedure. But, this isn't how it is really to be solved. There's a sleek way which I cannot find.
conic-sections
asked Jan 3 at 12:13
Ice InkberryIce Inkberry
336112
$endgroup$
The area of the triangle formed by the coordinate axes and tangent at vertex to the parabola whose focus is $(3,4)$ and tangents at $x=0$ and $y=0$ is?
I know how to do this, assume an equation for parabola with axis as $y=4/3 x$ and do all the procedure. But, this isn't how it is really to be solved. There's a sleek way which I cannot find.
conic-sections
conic-sections
asked Jan 3 at 12:13
Ice InkberryIce Inkberry
336112
asked Jan 3 at 12:13
Ice InkberryIce Inkberry
336112
asked Jan 3 at 12:13
Ice InkberryIce Inkberry
336112
asked Jan 3 at 12:13
Ice InkberryIce Inkberry
336112
asked Jan 3 at 12:13
Ice InkberryIce Inkberry
336112
336112
$begingroup$
This question is not well posed. There exist an infinite number of parabolas with focus at (3, 4). More information is needed. And "y= 4/3x" is NOT a parabola. A parabola with vertical line of symmetry can be written as $y= a(x- h)^2+ k$ and has focus $left(h, k+ frac{1}{4a}right)$. Saying that the focus is at (3, 4) means h= 3 and $k+ frac{1}{4a}= 4$ so $k= 4-frac{1}{4a}$. A parabola of the form $y= a(x- 3)^2+ 4- frac{1}{4a}$, for any a, has focus (3, 4).
$endgroup$
– user247327
Jan 3 at 12:19
$begingroup$
@user247327 No, the parabola has x=y=0 as tangents. Also, $y=4/3x$ is axis of the parabola.
$endgroup$
– Ice Inkberry
Jan 3 at 12:21
1
$begingroup$
How do you know that the parabola’s axis passes through the origin as you claim with that equation of the axis?
$endgroup$
– amd
Jan 3 at 21:46
add a comment |
$begingroup$
This question is not well posed. There exist an infinite number of parabolas with focus at (3, 4). More information is needed. And "y= 4/3x" is NOT a parabola. A parabola with vertical line of symmetry can be written as $y= a(x- h)^2+ k$ and has focus $left(h, k+ frac{1}{4a}right)$. Saying that the focus is at (3, 4) means h= 3 and $k+ frac{1}{4a}= 4$ so $k= 4-frac{1}{4a}$. A parabola of the form $y= a(x- 3)^2+ 4- frac{1}{4a}$, for any a, has focus (3, 4).
$endgroup$
– user247327
Jan 3 at 12:19
$begingroup$
@user247327 No, the parabola has x=y=0 as tangents. Also, $y=4/3x$ is axis of the parabola.
$endgroup$
– Ice Inkberry
Jan 3 at 12:21
1
$begingroup$
How do you know that the parabola’s axis passes through the origin as you claim with that equation of the axis?
$endgroup$
– amd
Jan 3 at 21:46
$begingroup$
This question is not well posed. There exist an infinite number of parabolas with focus at (3, 4). More information is needed. And "y= 4/3x" is NOT a parabola. A parabola with vertical line of symmetry can be written as $y= a(x- h)^2+ k$ and has focus $left(h, k+ frac{1}{4a}right)$. Saying that the focus is at (3, 4) means h= 3 and $k+ frac{1}{4a}= 4$ so $k= 4-frac{1}{4a}$. A parabola of the form $y= a(x- 3)^2+ 4- frac{1}{4a}$, for any a, has focus (3, 4).
$endgroup$
– user247327
Jan 3 at 12:19
$begingroup$
This question is not well posed. There exist an infinite number of parabolas with focus at (3, 4). More information is needed. And "y= 4/3x" is NOT a parabola. A parabola with vertical line of symmetry can be written as $y= a(x- h)^2+ k$ and has focus $left(h, k+ frac{1}{4a}right)$. Saying that the focus is at (3, 4) means h= 3 and $k+ frac{1}{4a}= 4$ so $k= 4-frac{1}{4a}$. A parabola of the form $y= a(x- 3)^2+ 4- frac{1}{4a}$, for any a, has focus (3, 4).
$endgroup$
– user247327
Jan 3 at 12:19
$begingroup$
@user247327 No, the parabola has x=y=0 as tangents. Also, $y=4/3x$ is axis of the parabola.
$endgroup$
– Ice Inkberry
Jan 3 at 12:21
$begingroup$
@user247327 No, the parabola has x=y=0 as tangents. Also, $y=4/3x$ is axis of the parabola.
$endgroup$
– Ice Inkberry
Jan 3 at 12:21
1
1
$begingroup$
How do you know that the parabola’s axis passes through the origin as you claim with that equation of the axis?
$endgroup$
– amd
Jan 3 at 21:46
$begingroup$
How do you know that the parabola’s axis passes through the origin as you claim with that equation of the axis?
$endgroup$
– amd
Jan 3 at 21:46
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
A nice property of the parabola states that: the perpendicular from the focus to any tangent intersects it, and the tangent through the vertex, at the same point.
Hence the tangent at the vertex intersects the axes at $(3,0)$ and $(0,4)$.
answered Jan 3 at 16:57
AretinoAretino
22.9k21443
$endgroup$
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%2f3060498%2fthe-area-of-the-triangle-formed-by-the-coordinate-axes-and-tangent-at-vertex-to%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
$begingroup$
A nice property of the parabola states that: the perpendicular from the focus to any tangent intersects it, and the tangent through the vertex, at the same point.
Hence the tangent at the vertex intersects the axes at $(3,0)$ and $(0,4)$.
answered Jan 3 at 16:57
AretinoAretino
22.9k21443
$endgroup$
add a comment |
$begingroup$
A nice property of the parabola states that: the perpendicular from the focus to any tangent intersects it, and the tangent through the vertex, at the same point.
Hence the tangent at the vertex intersects the axes at $(3,0)$ and $(0,4)$.
answered Jan 3 at 16:57
AretinoAretino
22.9k21443
$endgroup$
add a comment |
$begingroup$
A nice property of the parabola states that: the perpendicular from the focus to any tangent intersects it, and the tangent through the vertex, at the same point.
Hence the tangent at the vertex intersects the axes at $(3,0)$ and $(0,4)$.
answered Jan 3 at 16:57
AretinoAretino
22.9k21443
$endgroup$
A nice property of the parabola states that: the perpendicular from the focus to any tangent intersects it, and the tangent through the vertex, at the same point.
Hence the tangent at the vertex intersects the axes at $(3,0)$ and $(0,4)$.
answered Jan 3 at 16:57
AretinoAretino
22.9k21443
edited Jan 3 at 21:54
answered Jan 3 at 16:57
AretinoAretino
22.9k21443
answered Jan 3 at 16:57
AretinoAretino
22.9k21443
answered Jan 3 at 16:57
AretinoAretino
22.9k21443
22.9k21443
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%2f3060498%2fthe-area-of-the-triangle-formed-by-the-coordinate-axes-and-tangent-at-vertex-to%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
$begingroup$
This question is not well posed. There exist an infinite number of parabolas with focus at (3, 4). More information is needed. And "y= 4/3x" is NOT a parabola. A parabola with vertical line of symmetry can be written as $y= a(x- h)^2+ k$ and has focus $left(h, k+ frac{1}{4a}right)$. Saying that the focus is at (3, 4) means h= 3 and $k+ frac{1}{4a}= 4$ so $k= 4-frac{1}{4a}$. A parabola of the form $y= a(x- 3)^2+ 4- frac{1}{4a}$, for any a, has focus (3, 4).
$endgroup$
– user247327
Jan 3 at 12:19
$begingroup$
@user247327 No, the parabola has x=y=0 as tangents. Also, $y=4/3x$ is axis of the parabola.
$endgroup$
– Ice Inkberry
Jan 3 at 12:21
1
$begingroup$
How do you know that the parabola’s axis passes through the origin as you claim with that equation of the axis?
$endgroup$
– amd
Jan 3 at 21:46