MHB Finding focus, given four tangents

[Saitama]

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!

 

[Opalg]

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.

 

[Saitama]

Re: Finding focus, given four tangenets

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.

 

[Opalg]

Re: Finding focus, given four tangenets

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.

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 $q$.

 

[Saitama]

Re: Finding focus, given four tangenets

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?

Thank you!

 

[Opalg]

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 $(6/5,2/5)$.

 

[Saitama]

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.