Angle of the slope of an ellipse
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Find the relation between the angle made by a straight line connecting the origin and an ellipse (angle made between that line and the x axis) and the slope of the tangent to the ellipse. This should also give you the coordinates on the ellipse.
In particular at which point/points does the slope of the tangent is $45^{circ}$?
What is the angle the straight line connecting the origin to that point (points) make with the x axis? You can assume the equation of the ellipse is $$frac{x^{2}}{a^{2}}+frac{y^{2}}{b^{2}}=1,a>b$$.
For reference, for a circle of radius $a$, the slope of the tangent is 45 degrees at the points such that $y=pm x$. There are 4 such points one in each quadrant. The line connecting the center of the circle to those points also happens to make a 45 degree with the x axis. And the slope of the tangent at those points is also 45 degrees.
My approach would be to find the equation to the tangent of the ellipse, then find the equation of the line connecting the origin to the point $(x,y)$ of the ellipse and do the inner product of the two unit vectors along those two lines to find the angle between them. But then how to relate that to the angle of the line connecting the origin to the points in the ellipse? Or maybe there is an easier way to go about this?
conic-sections curves
asked Dec 8 '18 at 22:18
user35202user35202
31719
$endgroup$
add a comment |
$begingroup$
Find the relation between the angle made by a straight line connecting the origin and an ellipse (angle made between that line and the x axis) and the slope of the tangent to the ellipse. This should also give you the coordinates on the ellipse.
In particular at which point/points does the slope of the tangent is $45^{circ}$?
What is the angle the straight line connecting the origin to that point (points) make with the x axis? You can assume the equation of the ellipse is $$frac{x^{2}}{a^{2}}+frac{y^{2}}{b^{2}}=1,a>b$$.
For reference, for a circle of radius $a$, the slope of the tangent is 45 degrees at the points such that $y=pm x$. There are 4 such points one in each quadrant. The line connecting the center of the circle to those points also happens to make a 45 degree with the x axis. And the slope of the tangent at those points is also 45 degrees.
My approach would be to find the equation to the tangent of the ellipse, then find the equation of the line connecting the origin to the point $(x,y)$ of the ellipse and do the inner product of the two unit vectors along those two lines to find the angle between them. But then how to relate that to the angle of the line connecting the origin to the points in the ellipse? Or maybe there is an easier way to go about this?
conic-sections curves
asked Dec 8 '18 at 22:18
user35202user35202
31719
$endgroup$
$begingroup$
Am I missing something.. because you need to find the angle between the line through the origin and the tangent at that point. Both of which you have, so what's the problem?
$endgroup$
– mm-crj
Dec 8 '18 at 22:35
$begingroup$
Lets say the angle between the line connecting the origin and the point $(x,y)$ on the ellipse is $theta$. Lets says I know the tangent to the ellipse at that point $(x,y)$. How do I find the point(s) $(x,y)$ on the ellipse where the slope of the tangent is $45$ degrees and how do I find the angle $theta$ for that angle? More generally how do I find these for any slope angle?
$endgroup$
– user35202
Dec 8 '18 at 22:47
1
$begingroup$
First, find the point on the ellipse $(x_0 ,y_0 )$ for which the slope of the tangent$(dot{y})$ is $1$ and then find the slope of the line passing through that point and the origin. Does that help?
$endgroup$
– mm-crj
Dec 8 '18 at 22:50
1
$begingroup$
The slope to a point $(x,y)$ is $tanphi=ya^{2}/xb^{2}$. This is $1$ when $y/x=b^{2}/a^{2}$. The slope of the line through the origin and $(x,y)$ is $tantheta=y/x=b^{2}/a^{2}$. In general $tantheta=b^{2}/a^{2}tanphi$?
$endgroup$
– user35202
Dec 8 '18 at 23:26
1
$begingroup$
Ah I see the line can be either $pi/4$ 0r$3pi/4$ wrt the x asis
$endgroup$
– mm-crj
Dec 8 '18 at 23:50
add a comment |
$begingroup$
Find the relation between the angle made by a straight line connecting the origin and an ellipse (angle made between that line and the x axis) and the slope of the tangent to the ellipse. This should also give you the coordinates on the ellipse.
In particular at which point/points does the slope of the tangent is $45^{circ}$?
What is the angle the straight line connecting the origin to that point (points) make with the x axis? You can assume the equation of the ellipse is $$frac{x^{2}}{a^{2}}+frac{y^{2}}{b^{2}}=1,a>b$$.
For reference, for a circle of radius $a$, the slope of the tangent is 45 degrees at the points such that $y=pm x$. There are 4 such points one in each quadrant. The line connecting the center of the circle to those points also happens to make a 45 degree with the x axis. And the slope of the tangent at those points is also 45 degrees.
My approach would be to find the equation to the tangent of the ellipse, then find the equation of the line connecting the origin to the point $(x,y)$ of the ellipse and do the inner product of the two unit vectors along those two lines to find the angle between them. But then how to relate that to the angle of the line connecting the origin to the points in the ellipse? Or maybe there is an easier way to go about this?
conic-sections curves
asked Dec 8 '18 at 22:18
user35202user35202
31719
$endgroup$
Find the relation between the angle made by a straight line connecting the origin and an ellipse (angle made between that line and the x axis) and the slope of the tangent to the ellipse. This should also give you the coordinates on the ellipse.
In particular at which point/points does the slope of the tangent is $45^{circ}$?
What is the angle the straight line connecting the origin to that point (points) make with the x axis? You can assume the equation of the ellipse is $$frac{x^{2}}{a^{2}}+frac{y^{2}}{b^{2}}=1,a>b$$.
For reference, for a circle of radius $a$, the slope of the tangent is 45 degrees at the points such that $y=pm x$. There are 4 such points one in each quadrant. The line connecting the center of the circle to those points also happens to make a 45 degree with the x axis. And the slope of the tangent at those points is also 45 degrees.
My approach would be to find the equation to the tangent of the ellipse, then find the equation of the line connecting the origin to the point $(x,y)$ of the ellipse and do the inner product of the two unit vectors along those two lines to find the angle between them. But then how to relate that to the angle of the line connecting the origin to the points in the ellipse? Or maybe there is an easier way to go about this?
conic-sections curves
conic-sections curves
asked Dec 8 '18 at 22:18
user35202user35202
31719
asked Dec 8 '18 at 22:18
user35202user35202
31719
asked Dec 8 '18 at 22:18
user35202user35202
31719
asked Dec 8 '18 at 22:18
user35202user35202
31719
asked Dec 8 '18 at 22:18
user35202user35202
31719
31719
$begingroup$
Am I missing something.. because you need to find the angle between the line through the origin and the tangent at that point. Both of which you have, so what's the problem?
$endgroup$
– mm-crj
Dec 8 '18 at 22:35
$begingroup$
Lets say the angle between the line connecting the origin and the point $(x,y)$ on the ellipse is $theta$. Lets says I know the tangent to the ellipse at that point $(x,y)$. How do I find the point(s) $(x,y)$ on the ellipse where the slope of the tangent is $45$ degrees and how do I find the angle $theta$ for that angle? More generally how do I find these for any slope angle?
$endgroup$
– user35202
Dec 8 '18 at 22:47
1
$begingroup$
First, find the point on the ellipse $(x_0 ,y_0 )$ for which the slope of the tangent$(dot{y})$ is $1$ and then find the slope of the line passing through that point and the origin. Does that help?
$endgroup$
– mm-crj
Dec 8 '18 at 22:50
1
$begingroup$
The slope to a point $(x,y)$ is $tanphi=ya^{2}/xb^{2}$. This is $1$ when $y/x=b^{2}/a^{2}$. The slope of the line through the origin and $(x,y)$ is $tantheta=y/x=b^{2}/a^{2}$. In general $tantheta=b^{2}/a^{2}tanphi$?
$endgroup$
– user35202
Dec 8 '18 at 23:26
1
$begingroup$
Ah I see the line can be either $pi/4$ 0r$3pi/4$ wrt the x asis
$endgroup$
– mm-crj
Dec 8 '18 at 23:50
add a comment |
$begingroup$
Am I missing something.. because you need to find the angle between the line through the origin and the tangent at that point. Both of which you have, so what's the problem?
$endgroup$
– mm-crj
Dec 8 '18 at 22:35
$begingroup$
Lets say the angle between the line connecting the origin and the point $(x,y)$ on the ellipse is $theta$. Lets says I know the tangent to the ellipse at that point $(x,y)$. How do I find the point(s) $(x,y)$ on the ellipse where the slope of the tangent is $45$ degrees and how do I find the angle $theta$ for that angle? More generally how do I find these for any slope angle?
$endgroup$
– user35202
Dec 8 '18 at 22:47
1
$begingroup$
First, find the point on the ellipse $(x_0 ,y_0 )$ for which the slope of the tangent$(dot{y})$ is $1$ and then find the slope of the line passing through that point and the origin. Does that help?
$endgroup$
– mm-crj
Dec 8 '18 at 22:50
1
$begingroup$
The slope to a point $(x,y)$ is $tanphi=ya^{2}/xb^{2}$. This is $1$ when $y/x=b^{2}/a^{2}$. The slope of the line through the origin and $(x,y)$ is $tantheta=y/x=b^{2}/a^{2}$. In general $tantheta=b^{2}/a^{2}tanphi$?
$endgroup$
– user35202
Dec 8 '18 at 23:26
1
$begingroup$
Ah I see the line can be either $pi/4$ 0r$3pi/4$ wrt the x asis
$endgroup$
– mm-crj
Dec 8 '18 at 23:50
$begingroup$
Am I missing something.. because you need to find the angle between the line through the origin and the tangent at that point. Both of which you have, so what's the problem?
$endgroup$
– mm-crj
Dec 8 '18 at 22:35
$begingroup$
Am I missing something.. because you need to find the angle between the line through the origin and the tangent at that point. Both of which you have, so what's the problem?
$endgroup$
– mm-crj
Dec 8 '18 at 22:35
$begingroup$
Lets say the angle between the line connecting the origin and the point $(x,y)$ on the ellipse is $theta$. Lets says I know the tangent to the ellipse at that point $(x,y)$. How do I find the point(s) $(x,y)$ on the ellipse where the slope of the tangent is $45$ degrees and how do I find the angle $theta$ for that angle? More generally how do I find these for any slope angle?
$endgroup$
– user35202
Dec 8 '18 at 22:47
$begingroup$
Lets say the angle between the line connecting the origin and the point $(x,y)$ on the ellipse is $theta$. Lets says I know the tangent to the ellipse at that point $(x,y)$. How do I find the point(s) $(x,y)$ on the ellipse where the slope of the tangent is $45$ degrees and how do I find the angle $theta$ for that angle? More generally how do I find these for any slope angle?
$endgroup$
– user35202
Dec 8 '18 at 22:47
1
1
$begingroup$
First, find the point on the ellipse $(x_0 ,y_0 )$ for which the slope of the tangent$(dot{y})$ is $1$ and then find the slope of the line passing through that point and the origin. Does that help?
$endgroup$
– mm-crj
Dec 8 '18 at 22:50
$begingroup$
First, find the point on the ellipse $(x_0 ,y_0 )$ for which the slope of the tangent$(dot{y})$ is $1$ and then find the slope of the line passing through that point and the origin. Does that help?
$endgroup$
– mm-crj
Dec 8 '18 at 22:50
1
1
$begingroup$
The slope to a point $(x,y)$ is $tanphi=ya^{2}/xb^{2}$. This is $1$ when $y/x=b^{2}/a^{2}$. The slope of the line through the origin and $(x,y)$ is $tantheta=y/x=b^{2}/a^{2}$. In general $tantheta=b^{2}/a^{2}tanphi$?
$endgroup$
– user35202
Dec 8 '18 at 23:26
$begingroup$
The slope to a point $(x,y)$ is $tanphi=ya^{2}/xb^{2}$. This is $1$ when $y/x=b^{2}/a^{2}$. The slope of the line through the origin and $(x,y)$ is $tantheta=y/x=b^{2}/a^{2}$. In general $tantheta=b^{2}/a^{2}tanphi$?
$endgroup$
– user35202
Dec 8 '18 at 23:26
1
1
$begingroup$
Ah I see the line can be either $pi/4$ 0r$3pi/4$ wrt the x asis
$endgroup$
– mm-crj
Dec 8 '18 at 23:50
$begingroup$
Ah I see the line can be either $pi/4$ 0r$3pi/4$ wrt the x asis
$endgroup$
– mm-crj
Dec 8 '18 at 23:50
add a comment |
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$begingroup$
Am I missing something.. because you need to find the angle between the line through the origin and the tangent at that point. Both of which you have, so what's the problem?
$endgroup$
– mm-crj
Dec 8 '18 at 22:35
$begingroup$
Lets say the angle between the line connecting the origin and the point $(x,y)$ on the ellipse is $theta$. Lets says I know the tangent to the ellipse at that point $(x,y)$. How do I find the point(s) $(x,y)$ on the ellipse where the slope of the tangent is $45$ degrees and how do I find the angle $theta$ for that angle? More generally how do I find these for any slope angle?
$endgroup$
– user35202
Dec 8 '18 at 22:47
1
$begingroup$
First, find the point on the ellipse $(x_0 ,y_0 )$ for which the slope of the tangent$(dot{y})$ is $1$ and then find the slope of the line passing through that point and the origin. Does that help?
$endgroup$
– mm-crj
Dec 8 '18 at 22:50
1
$begingroup$
The slope to a point $(x,y)$ is $tanphi=ya^{2}/xb^{2}$. This is $1$ when $y/x=b^{2}/a^{2}$. The slope of the line through the origin and $(x,y)$ is $tantheta=y/x=b^{2}/a^{2}$. In general $tantheta=b^{2}/a^{2}tanphi$?
$endgroup$
– user35202
Dec 8 '18 at 23:26
1
$begingroup$
Ah I see the line can be either $pi/4$ 0r$3pi/4$ wrt the x asis
$endgroup$
– mm-crj
Dec 8 '18 at 23:50