Finding tangent points of an ellipse given an exterior point
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There are two tangents from the point $P = (frac{1261}{20},-1)$ to the ellipse with equation $frac{x^2}{169}+frac{y^2}{25} = 1.$
Determine the coordinates of the points, $E$ and $F$, where the tangents touch this ellipse. All coordinates should be correct to two decimal places.
So far I have found the slope to be $frac{-25x}{169y}$. Is it right to say the equation is $y+1 = frac{-25x}{169y}left(x-frac{1261}{20}right)$ ? I attempted to sub this equation of a line into the ellipse equation to get the points, but i still have two variables? All help appreciated.
geometry conic-sections
edited Nov 22 at 12:09
amWhy
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asked Nov 22 at 12:02
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up vote
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There are two tangents from the point $P = (frac{1261}{20},-1)$ to the ellipse with equation $frac{x^2}{169}+frac{y^2}{25} = 1.$
Determine the coordinates of the points, $E$ and $F$, where the tangents touch this ellipse. All coordinates should be correct to two decimal places.
So far I have found the slope to be $frac{-25x}{169y}$. Is it right to say the equation is $y+1 = frac{-25x}{169y}left(x-frac{1261}{20}right)$ ? I attempted to sub this equation of a line into the ellipse equation to get the points, but i still have two variables? All help appreciated.
geometry conic-sections
edited Nov 22 at 12:09
amWhy
191k28224439
asked Nov 22 at 12:02
Matlab rookie
307
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
There are two tangents from the point $P = (frac{1261}{20},-1)$ to the ellipse with equation $frac{x^2}{169}+frac{y^2}{25} = 1.$
Determine the coordinates of the points, $E$ and $F$, where the tangents touch this ellipse. All coordinates should be correct to two decimal places.
So far I have found the slope to be $frac{-25x}{169y}$. Is it right to say the equation is $y+1 = frac{-25x}{169y}left(x-frac{1261}{20}right)$ ? I attempted to sub this equation of a line into the ellipse equation to get the points, but i still have two variables? All help appreciated.
geometry conic-sections
edited Nov 22 at 12:09
amWhy
191k28224439
asked Nov 22 at 12:02
Matlab rookie
307
There are two tangents from the point $P = (frac{1261}{20},-1)$ to the ellipse with equation $frac{x^2}{169}+frac{y^2}{25} = 1.$
Determine the coordinates of the points, $E$ and $F$, where the tangents touch this ellipse. All coordinates should be correct to two decimal places.
So far I have found the slope to be $frac{-25x}{169y}$. Is it right to say the equation is $y+1 = frac{-25x}{169y}left(x-frac{1261}{20}right)$ ? I attempted to sub this equation of a line into the ellipse equation to get the points, but i still have two variables? All help appreciated.
geometry conic-sections
geometry conic-sections
edited Nov 22 at 12:09
amWhy
191k28224439
asked Nov 22 at 12:02
Matlab rookie
307
edited Nov 22 at 12:09
amWhy
191k28224439
asked Nov 22 at 12:02
Matlab rookie
307
edited Nov 22 at 12:09
amWhy
191k28224439
edited Nov 22 at 12:09
amWhy
191k28224439
edited Nov 22 at 12:09
amWhy
191k28224439
191k28224439
asked Nov 22 at 12:02
Matlab rookie
307
asked Nov 22 at 12:02
Matlab rookie
307
asked Nov 22 at 12:02
Matlab rookie
307
307
add a comment |
add a comment |
2 Answers
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Is it right to say the equation is $y+1 = frac{-25x}{169y}left(x-frac{1261}{20}right)$
No, that's not correct. When you say you've found the slope to be $frac{-25x}{169y}$, you need to be much more precise. What you know is that if a line is tangent to the ellipse at a point $(a, b)$ (which is on the ellipse, of course), then the slope of that line is $frac{-25a}{169b}$. That means the equation of the line is
$$
y - b = frac{-25a}{169b}(x-a).
$$
Notice that I did not use "$(x, y)$" as the point of tangency to the ellipse, because $(x, y)$ is used to denote a generic point of the tangent line, and only one of those points is on the ellipse.
So right there is where your approach went off the rails.
Let me suggest a different starting point:
Construct a line from $(1261/20, -1)$ to an arbitrary point $(a, b)$ on the ellipse. Determine the slope of the line, and compare this to the slope of the ellipse at $(a, b)$. For certain values of $(a, b)$, these two will be equal, and for those points, you've got a tangent line rather than a secant line.
answered Nov 22 at 12:21
John Hughes
62k24090
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0
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Calculate the polar's equation to
$$frac{frac{1261}{20}cdot x}{169}+frac{-1cdot y}{25}=1$$
first, solve for
$$y=25left(frac{97}{260}x-1right)$$
and plug that in the equation of the ellipse to get
$$x=frac{16}{5}quadtext{or}quad x=frac{312}{145}.$$
answered Nov 22 at 20:41
Michael Hoppe
10.7k31834
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2 Answers
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active
oldest
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2 Answers
2
active
oldest
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oldest
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active
oldest
votes
up vote
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down vote
accepted
Is it right to say the equation is $y+1 = frac{-25x}{169y}left(x-frac{1261}{20}right)$
No, that's not correct. When you say you've found the slope to be $frac{-25x}{169y}$, you need to be much more precise. What you know is that if a line is tangent to the ellipse at a point $(a, b)$ (which is on the ellipse, of course), then the slope of that line is $frac{-25a}{169b}$. That means the equation of the line is
$$
y - b = frac{-25a}{169b}(x-a).
$$
Notice that I did not use "$(x, y)$" as the point of tangency to the ellipse, because $(x, y)$ is used to denote a generic point of the tangent line, and only one of those points is on the ellipse.
So right there is where your approach went off the rails.
Let me suggest a different starting point:
Construct a line from $(1261/20, -1)$ to an arbitrary point $(a, b)$ on the ellipse. Determine the slope of the line, and compare this to the slope of the ellipse at $(a, b)$. For certain values of $(a, b)$, these two will be equal, and for those points, you've got a tangent line rather than a secant line.
answered Nov 22 at 12:21
John Hughes
62k24090
add a comment |
up vote
0
down vote
accepted
Is it right to say the equation is $y+1 = frac{-25x}{169y}left(x-frac{1261}{20}right)$
No, that's not correct. When you say you've found the slope to be $frac{-25x}{169y}$, you need to be much more precise. What you know is that if a line is tangent to the ellipse at a point $(a, b)$ (which is on the ellipse, of course), then the slope of that line is $frac{-25a}{169b}$. That means the equation of the line is
$$
y - b = frac{-25a}{169b}(x-a).
$$
Notice that I did not use "$(x, y)$" as the point of tangency to the ellipse, because $(x, y)$ is used to denote a generic point of the tangent line, and only one of those points is on the ellipse.
So right there is where your approach went off the rails.
Let me suggest a different starting point:
Construct a line from $(1261/20, -1)$ to an arbitrary point $(a, b)$ on the ellipse. Determine the slope of the line, and compare this to the slope of the ellipse at $(a, b)$. For certain values of $(a, b)$, these two will be equal, and for those points, you've got a tangent line rather than a secant line.
answered Nov 22 at 12:21
John Hughes
62k24090
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
Is it right to say the equation is $y+1 = frac{-25x}{169y}left(x-frac{1261}{20}right)$
No, that's not correct. When you say you've found the slope to be $frac{-25x}{169y}$, you need to be much more precise. What you know is that if a line is tangent to the ellipse at a point $(a, b)$ (which is on the ellipse, of course), then the slope of that line is $frac{-25a}{169b}$. That means the equation of the line is
$$
y - b = frac{-25a}{169b}(x-a).
$$
Notice that I did not use "$(x, y)$" as the point of tangency to the ellipse, because $(x, y)$ is used to denote a generic point of the tangent line, and only one of those points is on the ellipse.
So right there is where your approach went off the rails.
Let me suggest a different starting point:
Construct a line from $(1261/20, -1)$ to an arbitrary point $(a, b)$ on the ellipse. Determine the slope of the line, and compare this to the slope of the ellipse at $(a, b)$. For certain values of $(a, b)$, these two will be equal, and for those points, you've got a tangent line rather than a secant line.
answered Nov 22 at 12:21
John Hughes
62k24090
Is it right to say the equation is $y+1 = frac{-25x}{169y}left(x-frac{1261}{20}right)$
No, that's not correct. When you say you've found the slope to be $frac{-25x}{169y}$, you need to be much more precise. What you know is that if a line is tangent to the ellipse at a point $(a, b)$ (which is on the ellipse, of course), then the slope of that line is $frac{-25a}{169b}$. That means the equation of the line is
$$
y - b = frac{-25a}{169b}(x-a).
$$
Notice that I did not use "$(x, y)$" as the point of tangency to the ellipse, because $(x, y)$ is used to denote a generic point of the tangent line, and only one of those points is on the ellipse.
So right there is where your approach went off the rails.
Let me suggest a different starting point:
Construct a line from $(1261/20, -1)$ to an arbitrary point $(a, b)$ on the ellipse. Determine the slope of the line, and compare this to the slope of the ellipse at $(a, b)$. For certain values of $(a, b)$, these two will be equal, and for those points, you've got a tangent line rather than a secant line.
answered Nov 22 at 12:21
John Hughes
62k24090
answered Nov 22 at 12:21
John Hughes
62k24090
answered Nov 22 at 12:21
John Hughes
62k24090
answered Nov 22 at 12:21
John Hughes
62k24090
62k24090
add a comment |
add a comment |
up vote
0
down vote
Calculate the polar's equation to
$$frac{frac{1261}{20}cdot x}{169}+frac{-1cdot y}{25}=1$$
first, solve for
$$y=25left(frac{97}{260}x-1right)$$
and plug that in the equation of the ellipse to get
$$x=frac{16}{5}quadtext{or}quad x=frac{312}{145}.$$
answered Nov 22 at 20:41
Michael Hoppe
10.7k31834
add a comment |
up vote
0
down vote
Calculate the polar's equation to
$$frac{frac{1261}{20}cdot x}{169}+frac{-1cdot y}{25}=1$$
first, solve for
$$y=25left(frac{97}{260}x-1right)$$
and plug that in the equation of the ellipse to get
$$x=frac{16}{5}quadtext{or}quad x=frac{312}{145}.$$
answered Nov 22 at 20:41
Michael Hoppe
10.7k31834
add a comment |
up vote
0
down vote
up vote
0
down vote
Calculate the polar's equation to
$$frac{frac{1261}{20}cdot x}{169}+frac{-1cdot y}{25}=1$$
first, solve for
$$y=25left(frac{97}{260}x-1right)$$
and plug that in the equation of the ellipse to get
$$x=frac{16}{5}quadtext{or}quad x=frac{312}{145}.$$
answered Nov 22 at 20:41
Michael Hoppe
10.7k31834
Calculate the polar's equation to
$$frac{frac{1261}{20}cdot x}{169}+frac{-1cdot y}{25}=1$$
first, solve for
$$y=25left(frac{97}{260}x-1right)$$
and plug that in the equation of the ellipse to get
$$x=frac{16}{5}quadtext{or}quad x=frac{312}{145}.$$
answered Nov 22 at 20:41
Michael Hoppe
10.7k31834
answered Nov 22 at 20:41
Michael Hoppe
10.7k31834
answered Nov 22 at 20:41
Michael Hoppe
10.7k31834
answered Nov 22 at 20:41
Michael Hoppe
10.7k31834
10.7k31834
add a comment |
add a comment |
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