Finding focus, given two tangents

[Saitama]

Problem:

Let focus of parabola is $S(x,y)$ which touches the x-axis and $y=x$ at $(1,0)$ and $(1,1)$ respectively, then find the focus of parabola.

Attempt:
Honestly, I am lost on this. I don't know how to make use of the given information to find the focus. I need a few hints to start with this problem. :(

Any help is appreciated. Thanks!

 

[Opalg]

I don't know of an easy way to do this. I started by finding the equation of the parabola. The general parabola has equation of the form $$(ax+by)^2 = cx+dy+e.$$ Multiplying the equation through by a constant does not alter it, so we may as well assume that $a=1$: $$(x+by)^2 = cx+dy+e.$$ We know that the line $y=0$ touches the parabola where $x=1$. So if we put $y=0$ in the equation of the parabola it should reduce to $(x-1)^2=0$. Similarly, the line $y=x$ touches the parabola where $x=1$. So if we put $y=x$ in the equation of the parabola it should again reduce to (x-1)^2=0$. That gives you enough information to deduce the equation of the parabola.
[sp]$(x-2y)^2 = 2x-1$.[/sp]
To find the focus, I used the property that light rays coming from infinity, reflected in the parabola, converge at the focus. From the equation of the parabola, you should see that these light rays will come from infinity along a line with gradient $1/2$. So the ray meeting the parabola at the point $(1,0)$ will bounce off the horizontal tangent there along a line with slope $-1/2$. Therefore the focus must lie on the line $y = -\frac12(x-1)$. The calculation at the other point of tangency is more complicated because the tangent there has slope $1$. So you need to do some work to find the slope of the reflected line. Once you have done that, it is easy to find the focus as the point where the two reflected lines meet.
[sp]The other reflected line has equation $y=2x-1$. The focus is at $\bigl(\frac35,\frac15\bigr)$.[/sp]

 

[Saitama]

Hi Opalg! :)

I appreciate the time you spent to solve this problem and explain the steps but I am stuck at the beginning statements. (Doh)

So if we put $y=0$ in the equation of the parabola it should reduce to $(x-1)^2=0$.

If I put y=0, I get $x^2=e$.

I am guessing that I have to make use of the point of contact $(1,0)$ and get the equation $(x-1)^2=0$ but I don't see how. I am honestly lost. :(

Thank you!

 

[Opalg]

Sorry, I am away from home and only have a phone connection for the next three weeks. Maybe someone else can help you.

 

[I like Serena]

If I put y=0, I get $x^2=e$.

I am guessing that I have to make use of the point of contact $(1,0)$ and get the equation $(x-1)^2=0$ but I don't see how. I am honestly lost.

I did something similar.
Fill in the 2 points into the equation.
Take the derivative of the equation and fill in those 2 points again.
Combine those 4 equations and you should be able to deduce all coefficients.

 

[Saitama]

I did something similar.
Fill in the 2 points into the equation.
Take the derivative of the equation and fill in those 2 points again.
Combine those 4 equations and you should be able to deduce all coefficients.

Hi ILS! :)

I tried the following:

$$(x+by)^2=cx+dy+e\,\,\,\, (*)$$
and differentiating
$$2(x+by)(1+by')=c+dy'\,\,\,\, (**)$$

Substituting $y=0$ and $x=1$ in $(*)$, I get $\fbox{c+e=1}$. Substituting $y=1$ and $x=1$ in the same equation, I get $(1+b)^2=c+d+e=d+1\,\,\,\, (***)$.

With $x=1,y=0$ and $y'=0$ in $(**)$, I get $\fbox{c=2}$. Hence $\fbox{e=-1}$.
Similarly, using $y=1,x=1$ and $y'=1$, I have, $2(1+b)^2=c+d$. From $(***)$, $d+2=2(d+1) \Rightarrow \fbox{d=0}$. Hence, $\fbox{b=-2 or 0}$.

I have got all the coefficients but I am unsure about which value to pick for $b$. Using $b=0$ gives an equation with no $y$ so I think I have to use the $b=2$. I then get the equation shown by Opalg i.e $(x-2y)^2=2x-1$.

I cannot understand why do I need to find the other reflected line? Won't the focus be simply the intersection of $2y=x$ (axis of parabola) and one of the reflected lines Opalg found?

I do understand what I am suggesting is incorrect because finding the intersection gives an incorrect answer but I don't see why it is wrong to do so.

Thanks!

 

[Opalg]

Axis is not $2y=x $ but the parallel line $2y=x-\frac15$.

 

[MarkFL]

The parabola's axis of symmetry does not pass through the origin. I see Opalg has already informed you of this. :D

Here is an outline of a rotation of axes method:

Spoiler

Knowing the slope of this axis is \(\displaystyle \frac{1}{2}\), we may rotate the axes appropriately:

\(\displaystyle x=\frac{2}{\sqrt{5}}X-\frac{1}{\sqrt{5}}Y\)

\(\displaystyle y=\frac{1}{\sqrt{5}}X+\frac{2}{\sqrt{5}}Y\)

And then the parabola becomes:

\(\displaystyle 5Y^2=\frac{4}{\sqrt{5}}X-\frac{2}{\sqrt{5}}Y-1\)

Arranged in vertex form, this is:

\(\displaystyle X=\frac{5\sqrt{5}}{4}\left(Y+\frac{1}{5\sqrt{5}} \right)^2+\frac{6}{5\sqrt{5}}\)

From this we obtain the focus:

\(\displaystyle \left(\frac{7}{5\sqrt{5}},-\frac{1}{5\sqrt{5}} \right)\)

Mapping these coordinates back into the original coordinate system, we find the focus at:

\(\displaystyle \left(\frac{3}{5},\frac{1}{5} \right)\)

 

[Saitama]

The parabola's axis of symmetry does not pass through the origin. I see Opalg has already informed you of this. :D

Here is an outline of a rotation of axes method:

Spoiler

Knowing the slope of this axis is \(\displaystyle \frac{1}{2}\), we may rotate the axes appropriately:

\(\displaystyle x=\frac{2}{\sqrt{5}}X-\frac{1}{\sqrt{5}}Y\)

\(\displaystyle y=\frac{1}{\sqrt{5}}X+\frac{2}{\sqrt{5}}Y\)

And then the parabola becomes:

\(\displaystyle 5Y^2=\frac{4}{\sqrt{5}}X-\frac{2}{\sqrt{5}}Y-1\)

Arranged in vertex form, this is:

\(\displaystyle X=\frac{5\sqrt{5}}{4}\left(Y+\frac{1}{5\sqrt{5}} \right)^2+\frac{6}{5\sqrt{5}}\)

From this we obtain the focus:

\(\displaystyle \left(\frac{7}{5\sqrt{5}},-\frac{1}{5\sqrt{5}} \right)\)

Mapping these coordinates back into the original coordinate system, we find the focus at:

\(\displaystyle \left(\frac{3}{5},\frac{1}{5} \right)\)

Axis is not $2y=x $ but the parallel line $2y=x-\frac15$.

I was digging through my notes and I found something. My teacher taught a second way to interpret an equation of parabola which is as follows:

$$(\text{Perpendicular distance from axis})^2=(\text{Length of latus rectum})\times (\text{Perpendicular distance from tangent at vertex})$$

And then we did a problem on finding the axis of a weird parabola. I apply the same method my teacher taught for that problem.

Rewrite the equation as
$$(x-2y+\lambda)^2=2x-1+\lambda^2+2x\lambda-2y\lambda$$
$$\Rightarrow (x-2y+\lambda)^2=2x(\lambda+1)-2\lambda y+\lambda^2-1$$

The left side of the equation represents the square of axis and the right side, the tangent at vertex. (I am not sure why this is so as its been a long time I did this kind of problem, can someone please clarify this? Thanks! )

The product of slopes must be -1 and using that I get, $\lambda=-1/3$. But I should get $\lambda=-1/5$ , I don't get why this method doesn't work. (Sweating)

 

[I like Serena]

I was digging through my notes and I found something. My teacher taught a second way to interpret an equation of parabola which is as follows:

$$(\text{Perpendicular distance from axis})^2=(\text{Length of latus rectum})\times (\text{Perpendicular distance from tangent at vertex})$$

And then we did a problem on finding the axis of a weird parabola. I apply the same method my teacher taught for that problem.

Rewrite the equation as
$$(x-2y+\lambda)^2=2x-1+\lambda^2+2x\lambda-2y\lambda$$
$$\Rightarrow (x-2y+\lambda)^2=2x(\lambda+1)-2\lambda y+\lambda^2-1$$

The left side of the equation represents the square of axis and the right side, the tangent at vertex. (I am not sure why this is so as its been a long time I did this kind of problem, can someone please clarify this? Thanks! )

The product of slopes must be -1 and using that I get, $\lambda=-1/3$. But I should get $\lambda=-1/5$ , I don't get why this method doesn't work. (Sweating)

The general form of a parabola can be expressed as:
$$y^2 = 4ax$$
where $a$ is the distance of the focus point to the vertex, which is also the distance of the vertex to the directrix. Furthermore, the semi-latus rectum has length $\ell = 2a$.
See figure.

We can rewrite this with:
\begin{array}{}
y &=& \text{Perpendicular distance to axis of symmetry} \\
x &=& \text{Perpendicular distance to tangent at vertex} \\
\ell = 2a &=& \text{Length of semi-latus rectum}
\end{array}
as
\begin{array}{}
(\text{Perpendicular distance to axis of symmetry})^2 = 2 &\times& \text{Length of semi-latus rectum} \\
&\times& \text{Perpendicular distance to tangent at vertex}
\end{array}

Now let's rotate the parabola by an angle $\phi$ and translate it by $(p,q)$.
And let's define $c = \cos \phi, \ s = \sin \phi$.
Then the equation becomes:
$$(cX - sY - q)^2 = 4a (sX + cY - p)$$

As you can see, this is a close match to your equation.
If you can bring the equation in this form, you can read off all relevant parameters immediately.

 

[Saitama]

The general form of a parabola can be expressed as:
$$y^2 = 4ax$$
where $a$ is the distance of the focus point to the vertex, which is also the distance of the vertex to the directrix. Furthermore, the latus rectum has length $\ell = 2a$.
See figure.

We can rewrite this with:
\begin{array}{}
y &=& \text{Perpendicular distance to axis of symmetry} \\
x &=& \text{Perpendicular distance to tangent at vertex} \\
\ell = 2a &=& \text{Length of latus rectum}
\end{array}
as
\begin{array}{}
(\text{Perpendicular distance to axis of symmetry})^2 = 2 &\times& \text{Length of latus rectum} \\
&\times& \text{Perpendicular distance to tangent at vertex}
\end{array}

Length of latus rectum is $4a$. :)

Latus Rectum -- from Wolfram MathWorld

I found the error in my working. I wrote:
$$(x-2y+\lambda)^2=2x-1+\lambda^2+2x\lambda-2y\lambda$$

...but it should have been this:

$$\begin{array} {}
(x-2y+\lambda)^2=2x-1+\lambda^2+2x\lambda-4y\lambda \\
\Rightarrow (x-2y+\lambda)^2=2x(\lambda+1)-4\lambda y+\lambda^2-1\\
\end{array}$$
Equating the product of slopes to -1, I have
$$\frac{1}{2}\frac{2(\lambda+1)}{4\lambda}=-1 \Rightarrow \lambda=-\frac{1}{5}$$
Substituting this value of $\lambda$, I get
$$(x-2y-1/5)^2=\frac{4}{5}(2x+y-6)$$
Writing in terms of perpendicular distances, I have:
$$\left(\frac{x-2y-1/5}{\sqrt{5}}\right)^2=\frac{4}{5\sqrt{5}}\left( \frac{2x+y-6}{\sqrt{5}} \right)$$
Hence, the axis of parabola is $x-2y-1/5=0$ and therefore, the focus is $(3/5,1/5)$.

Thanks a lot Opalg and ILS! :)

I have learned quite a lot from this thread. Thanks again! :D

 

[I like Serena]

Length of latus rectum is $4a$. :)

Latus Rectum -- from Wolfram MathWorld

Oh!
I always thought $\ell$ was the latus rectum.
It's only now that I'm learning that $\ell$ is actually the semi-latus rectum.