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.






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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







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      up vote
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      favorite









      up vote
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      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







      share|cite|improve this question















      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






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      edited Nov 22 at 12:09









      amWhy

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      asked Nov 22 at 12:02









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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.






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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
            $$x=frac{16}{5}quadtext{or}quad x=frac{312}{145}.$$






            share|cite|improve this answer





















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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.






              share|cite|improve this answer

























                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.






                share|cite|improve this answer























                  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.






                  share|cite|improve this answer













                  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.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 22 at 12:21









                  John Hughes

                  62k24090




                  62k24090






















                      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
                      $$x=frac{16}{5}quadtext{or}quad x=frac{312}{145}.$$






                      share|cite|improve this answer

























                        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
                        $$x=frac{16}{5}quadtext{or}quad x=frac{312}{145}.$$






                        share|cite|improve this answer























                          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
                          $$x=frac{16}{5}quadtext{or}quad x=frac{312}{145}.$$






                          share|cite|improve this answer












                          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
                          $$x=frac{16}{5}quadtext{or}quad x=frac{312}{145}.$$







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Nov 22 at 20:41









                          Michael Hoppe

                          10.7k31834




                          10.7k31834






























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