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Finding the equation of a parabola, given the length of a portion of a focal chord, and the angle the chord makes with the parabola's axis

1

$begingroup$

Find the equation of the parabola on a picture if $|FL|=8$ units and $angle KFO=60^o$. $F$ is given as the focus of the parabola.

We know that this parabola passes through the point $(0,0)$, so if I can find a different point lying on this parabola I can find its equation. But I can't find this point.

How can I solve this problem?

geometry analytic-geometry conic-sections

share|cite|improve this question

edited Mar 23 at 13:25

Maria Mazur

50k1361125

asked Mar 23 at 8:19

Eldar RahimliEldar Rahimli

44410

$endgroup$

add a comment | 

1

$begingroup$

Find the equation of the parabola on a picture if $|FL|=8$ units and $angle KFO=60^o$. $F$ is given as the focus of the parabola.

We know that this parabola passes through the point $(0,0)$, so if I can find a different point lying on this parabola I can find its equation. But I can't find this point.

How can I solve this problem?

geometry analytic-geometry conic-sections

share|cite|improve this question

edited Mar 23 at 13:25

Maria Mazur

50k1361125

asked Mar 23 at 8:19

Eldar RahimliEldar Rahimli

44410

$endgroup$

add a comment | 

$begingroup$

Find the equation of the parabola on a picture if $|FL|=8$ units and $angle KFO=60^o$. $F$ is given as the focus of the parabola.

We know that this parabola passes through the point $(0,0)$, so if I can find a different point lying on this parabola I can find its equation. But I can't find this point.

How can I solve this problem?

geometry analytic-geometry conic-sections

share|cite|improve this question

edited Mar 23 at 13:25

Maria Mazur

50k1361125

asked Mar 23 at 8:19

Eldar RahimliEldar Rahimli

44410

$endgroup$

share|cite|improve this question

edited Mar 23 at 13:25

Maria Mazur

50k1361125

asked Mar 23 at 8:19

Eldar RahimliEldar Rahimli

44410

share|cite|improve this question

edited Mar 23 at 13:25

Maria Mazur

50k1361125

asked Mar 23 at 8:19

Eldar RahimliEldar Rahimli

44410

edited Mar 23 at 13:25

Maria Mazur

50k1361125

edited Mar 23 at 13:25

Maria Mazur

50k1361125

asked Mar 23 at 8:19

Eldar RahimliEldar Rahimli

44410

asked Mar 23 at 8:19

Eldar RahimliEldar Rahimli

44410

1 Answer
1

active

oldest

votes

5

$begingroup$

Guide: The equation of this parabola is $y^2=2px$ for some real (and negative) $p$. Then $F=({pover 2},0)$ and a line $d$ has an equation $$y=sqrt{3}left(x-{pover 2}right)$$

Solving the equation $$3left(x-{pover 2}right)^2 = 2px$$you will get $x$ for $L$ (and $K$) and then you can calculate the $y$ for $L$. Use the fact $LF =8$ and the formula for the distance between two points and you will get a $p$.

share|cite|improve this answer

edited Mar 23 at 12:10

Rory Daulton

29.6k63355

answered Mar 23 at 8:28

Maria MazurMaria Mazur

50k1361125

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  I like your answer, +1 from me--you do not give all details, which is appropriate for a question that shows no detailed work from the questioner. I could quibble a bit--I would use $f$, the distance from point $F$ to the origin, rather than $p$ in the equations, since it has a direct meaning in the diagram and avoids some fractions in the equations. I'll also edit your parentheses to make them look slightly better. But great job!
  $endgroup$
  – Rory Daulton
  Mar 23 at 12:09
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @RoryDaulton,I'm curious how am I supposed to show my non-existent detailed work on a problem which I have no idea how to solve, hence I have posted it here!? I really tried to think it through, indeed it was the third time I tried, but I am absolutely stumped.
  $endgroup$
  – Eldar Rahimli
  Mar 23 at 16:42
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @EldarRahimli: You did show some work, so I did not downvote or vote to close your question. But now Maria Mazur has given you an excellent start and pointed the way to finish the problem. If you do not understand this answer, leave a comment here saying which part you do not understand and Maria or others can help more. If you do understand, do the work that Maria suggested. If you get stuck on that, edit your question to show how far you got and ask for more help. In short, you could show some details now then ask for more details in an answer.
  $endgroup$
  – Rory Daulton
  Mar 23 at 17:03
  

  
  
  
  

add a comment | 

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5

$begingroup$

Guide: The equation of this parabola is $y^2=2px$ for some real (and negative) $p$. Then $F=({pover 2},0)$ and a line $d$ has an equation $$y=sqrt{3}left(x-{pover 2}right)$$

Solving the equation $$3left(x-{pover 2}right)^2 = 2px$$you will get $x$ for $L$ (and $K$) and then you can calculate the $y$ for $L$. Use the fact $LF =8$ and the formula for the distance between two points and you will get a $p$.

share|cite|improve this answer

edited Mar 23 at 12:10

Rory Daulton

29.6k63355

answered Mar 23 at 8:28

Maria MazurMaria Mazur

50k1361125

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  I like your answer, +1 from me--you do not give all details, which is appropriate for a question that shows no detailed work from the questioner. I could quibble a bit--I would use $f$, the distance from point $F$ to the origin, rather than $p$ in the equations, since it has a direct meaning in the diagram and avoids some fractions in the equations. I'll also edit your parentheses to make them look slightly better. But great job!
  $endgroup$
  – Rory Daulton
  Mar 23 at 12:09
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @RoryDaulton,I'm curious how am I supposed to show my non-existent detailed work on a problem which I have no idea how to solve, hence I have posted it here!? I really tried to think it through, indeed it was the third time I tried, but I am absolutely stumped.
  $endgroup$
  – Eldar Rahimli
  Mar 23 at 16:42
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @EldarRahimli: You did show some work, so I did not downvote or vote to close your question. But now Maria Mazur has given you an excellent start and pointed the way to finish the problem. If you do not understand this answer, leave a comment here saying which part you do not understand and Maria or others can help more. If you do understand, do the work that Maria suggested. If you get stuck on that, edit your question to show how far you got and ask for more help. In short, you could show some details now then ask for more details in an answer.
  $endgroup$
  – Rory Daulton
  Mar 23 at 17:03
  

  
  
  
  

add a comment | 

5

$begingroup$

Guide: The equation of this parabola is $y^2=2px$ for some real (and negative) $p$. Then $F=({pover 2},0)$ and a line $d$ has an equation $$y=sqrt{3}left(x-{pover 2}right)$$

Solving the equation $$3left(x-{pover 2}right)^2 = 2px$$you will get $x$ for $L$ (and $K$) and then you can calculate the $y$ for $L$. Use the fact $LF =8$ and the formula for the distance between two points and you will get a $p$.

share|cite|improve this answer

edited Mar 23 at 12:10

Rory Daulton

29.6k63355

answered Mar 23 at 8:28

Maria MazurMaria Mazur

50k1361125

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  I like your answer, +1 from me--you do not give all details, which is appropriate for a question that shows no detailed work from the questioner. I could quibble a bit--I would use $f$, the distance from point $F$ to the origin, rather than $p$ in the equations, since it has a direct meaning in the diagram and avoids some fractions in the equations. I'll also edit your parentheses to make them look slightly better. But great job!
  $endgroup$
  – Rory Daulton
  Mar 23 at 12:09
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @RoryDaulton,I'm curious how am I supposed to show my non-existent detailed work on a problem which I have no idea how to solve, hence I have posted it here!? I really tried to think it through, indeed it was the third time I tried, but I am absolutely stumped.
  $endgroup$
  – Eldar Rahimli
  Mar 23 at 16:42
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @EldarRahimli: You did show some work, so I did not downvote or vote to close your question. But now Maria Mazur has given you an excellent start and pointed the way to finish the problem. If you do not understand this answer, leave a comment here saying which part you do not understand and Maria or others can help more. If you do understand, do the work that Maria suggested. If you get stuck on that, edit your question to show how far you got and ask for more help. In short, you could show some details now then ask for more details in an answer.
  $endgroup$
  – Rory Daulton
  Mar 23 at 17:03
  

  
  
  
  

add a comment | 

$begingroup$

Guide: The equation of this parabola is $y^2=2px$ for some real (and negative) $p$. Then $F=({pover 2},0)$ and a line $d$ has an equation $$y=sqrt{3}left(x-{pover 2}right)$$

Solving the equation $$3left(x-{pover 2}right)^2 = 2px$$you will get $x$ for $L$ (and $K$) and then you can calculate the $y$ for $L$. Use the fact $LF =8$ and the formula for the distance between two points and you will get a $p$.

share|cite|improve this answer

edited Mar 23 at 12:10

Rory Daulton

29.6k63355

answered Mar 23 at 8:28

Maria MazurMaria Mazur

50k1361125

$endgroup$

share|cite|improve this answer

edited Mar 23 at 12:10

Rory Daulton

29.6k63355

answered Mar 23 at 8:28

Maria MazurMaria Mazur

50k1361125

edited Mar 23 at 12:10

Rory Daulton

29.6k63355

edited Mar 23 at 12:10

Rory Daulton

29.6k63355

answered Mar 23 at 8:28

Maria MazurMaria Mazur

50k1361125

answered Mar 23 at 8:28

Maria MazurMaria Mazur

50k1361125