Finding focus, given four tangents

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In summary, we are trying to find the coordinates of the focus of a parabola given two pairs of perpendicular tangents. By setting up equations for the directrix and using the fact that the distance from a point on the parabola to a line is equal to its distance from the focus, we can find the equation of the parabola. Then, by setting the discriminant of this equation to be equal to zero, we can solve for the coordinates of the focus. Alternatively, we can use a geometric approach by finding the equations of the dashed circles formed by the tangents and finding the intersection points. Both methods yield the same solution, $(6/5, 2/5)$.
  • #1
Saitama
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Problem:
Find the coordinates of focus of a parabola which $x=0, y=0, x+y=1$ and $y=x-2$.

Attempt:
I noticed that the we are given two pairs of perpendicular tangents. Since perpendicular tangents intersect at directrix, the equation of directrix is $y=-x/3$.

I am clueless about the next step.

Any help is appreciated. Thanks!
 
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  • #2
Re: Finding focus, given four tangenets

That is a very good start. Suppose the focus is at the point $(p,q)$. Then the parabola consists of points whose distance from $(p,q)$ is the same as their distance from the line $x+3y=0.$ The square of the distance from the point $(x,y)$ to that line is $\dfrac{(x+3y)^2}{1^2+3^2}.$ So the equation of the parabola is $\dfrac{(x+3y)^2}{10} = (x-p)^2 + (y-q)^2.$

Having got that far, you then need to find $p$ and $q$ from the fact that the four given lines are tangents to the parabola.
 
  • #3
Re: Finding focus, given four tangenets

Opalg said:
Having got that far, you then need to find $p$ and $q$ from the fact that the four given lines are tangents to the parabola.

I am not sure but don't we need the point of contacts to use the above condition? When the equation of parabola is differentiated, the DE consists of $x$, $y$ and $y'$, I know $y'$ but not the other two. :confused:
 
  • #4
Re: Finding focus, given four tangenets

Pranav said:
I am not sure but don't we need the point of contacts to use the above condition? When the equation of parabola is differentiated, the DE consists of $x$, $y$ and $y'$, I know $y'$ but not the other two. :confused:
I think it is easier to do this by another method. You have a parabola $(x+3y)^2 = 10(x-p)^2 + 10(x-q)^2$ and four lines, one of which is the coordinate axis $y=0$. If you put $y=0$ in the equation of the parabola then it becomes $x^2 = 10(x-p)^2 + 10q^2$, or $9x^2 - 20px + 10(p^2+q^2) = 0$. That is a quadratic equation in $x$, whose solutions are the $x$-coordinates of the points where the line meets the parabola. A quadratic equation can have either two solutions, or no real solutions, or one repeated solution, depending on whether its discriminant "$b^2 - 4ac$" is positive, negative or zero. If that line is going to be a tangent to the parabola then it will meet it in just one, repeated, point. So the condition for the line to be a tangent is that the discriminant of that quadratic should be zero. In other words, $(20p)^2 - 4\cdot9 \cdot 10(p^2+q^2) = 0$. That gives you an equation connecting $p$ and $q$. Now do the same thing for another of those tangent lines (preferably not the other coordinate axis $x=0$, which would just give you the same equation again), and you will have a second equation for $p$ and $q$.
 
  • #5
Re: Finding focus, given four tangenets

Opalg said:
I think it is easier to do this by another method. You have a parabola $(x+3y)^2 = 10(x-p)^2 + 10(x-q)^2$ and four lines, one of which is the coordinate axis $y=0$. If you put $y=0$ in the equation of the parabola then it becomes $x^2 = 10(x-p)^2 + 10q^2$, or $9x^2 - 20px + 10(p^2+q^2) = 0$. That is a quadratic equation in $x$, whose solutions are the $x$-coordinates of the points where the line meets the parabola. A quadratic equation can have either two solutions, or no real solutions, or one repeated solution, depending on whether its discriminant "$b^2 - 4ac$" is positive, negative or zero. If that line is going to be a tangent to the parabola then it will meet it in just one, repeated, point. So the condition for the line to be a tangent is that the discriminant of that quadratic should be zero. In other words, $(20p)^2 - 4\cdot9 \cdot 10(p^2+q^2) = 0$. That gives you an equation connecting $p$ and $q$. Now do the same thing for another of those tangent lines (preferably not the other coordinate axis $x=0$, which would just give you the same equation again), and you will have a second equation for $p$ and $q$.

Awesome! Thanks a lot Opalg! :D

I used the other line $x+y=1$ as you said and got the given answer i.e (6/5,2/5).

Now that I have reached the answer, I compared our solution with the given solution. Please look at the attachment.

I have no idea what the solution is doing, can you please help me understand the solution? :confused:

Thank you!
 

Attachments

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    solution of parabola.png
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  • #6
I agree with your solution $(6/5,2/5)$. That geometric solution is extremely slick. It relies on a theorem (unknown to me) saying that the circumcircle of the triangle formed by any three tangents to a parabola passes through the focus. If you accept that theorem then the next step is to find the equation of the dashed circles in the diagram. [There is a misprint in the diagram: the origin should be labelled as $(0,0)$, not $(0,1)$.] The smaller circle passes through $(0,0)$, $(1,0)$ and $(0,1)$. You can quickly find its equation using the fact that the equation of a circle is always of the form $x^2 + y^2 + ax + by + c = 0$. If it goes through the origin then $c=0$. Then if it goes through $(1,0)$ you must have $a=-1$ and if it goes through $(0,1)$ you must have $b=-1$. So its equation is $x(x-1) + y(y-1) = 0$. Similarly for the larger circle.

The two circles intersect at the points where $x(x-1) + y(y-1) = 0$ and $x(x-2) + y(y+2) = 0$. Solving those equations (start by subtracting one from the other), you find that the intersection points are $(0,0)$ and $(6/5,2/5)$.
 
  • #7
Opalg said:
I agree with your solution $(6/5,2/5)$. That geometric solution is extremely slick. It relies on a theorem (unknown to me) saying that the circumcircle of the triangle formed by any three tangents to a parabola passes through the focus. If you accept that theorem then the next step is to find the equation of the dashed circles in the diagram. [There is a misprint in the diagram: the origin should be labelled as $(0,0)$, not $(0,1)$.] The smaller circle passes through $(0,0)$, $(1,0)$ and $(0,1)$. You can quickly find its equation using the fact that the equation of a circle is always of the form $x^2 + y^2 + ax + by + c = 0$. If it goes through the origin then $c=0$. Then if it goes through $(1,0)$ you must have $a=-1$ and if it goes through $(0,1)$ you must have $b=-1$. So its equation is $x(x-1) + y(y-1) = 0$. Similarly for the larger circle.

The two circles intersect at the points where $x(x-1) + y(y-1) = 0$ and $x(x-2) + y(y+2) = 0$. Solving those equations (start by subtracting one from the other), you find that the intersection points are $(0,0)$ and $(6/5,2/5)$.

Thanks Opalg once again! :)

Btw, I found something which could be of interest. Here it is: The Parabola . Scroll down to Lambert's theorem.
 

FAQ: Finding focus, given four tangents

1. What is the concept of "finding focus, given four tangents"?

The concept of "finding focus, given four tangents" involves determining the location of a point or object that is equidistant from four given points or objects, known as tangents. This concept is commonly used in mathematics and geometry, as well as in other scientific fields.

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4. Are there any limitations to using "finding focus, given four tangents"?

While "finding focus, given four tangents" is a useful concept in many scientific applications, it does have limitations. For example, it can only be used to determine the location of a point if the four tangents are not all parallel. Additionally, it may not be applicable in situations where the tangents do not accurately represent the desired point or object.

5. How can "finding focus, given four tangents" be solved mathematically?

The mathematical method for solving "finding focus, given four tangents" involves using the properties of circles and tangents to create a system of equations and solve for the unknown point. This can be done using algebraic or geometric methods, depending on the given information and desired outcome.

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