# Hyperbola Help

#### Nathan0707

##### New member
Hi, could someone please help me solve this question using calculus and non-calculus. If someone could walk me through showing steps I would be very thankful. Help is be much appreciated!

#### MarkFL

##### Super Moderator
Staff member
Hello, and welcome to FMH!

I would begin by writing the coordinates of the foci in terms of the parameters of the given hyperbolic equation. Can you state these?

#### Nathan0707

##### New member
I believe I was able to solve using calculus.
I did: Der(x^2/a^2)-Der(y^2/b^2)=Der(1)
Taking that equation I did the same but with the x0 and y0 for the points of the center: (x^2+xo)/a^2-(y^2+yo)/b^2=1
I then solved for y: y=b/a(x+xo)+yo
And this shows that beta and alpha are a ratio of each other.
Would this be correct?

#### Jomo

##### Elite Member
I believe I was able to solve using calculus.
I did: Der(x^2/a^2)-Der(y^2/b^2)=Der(1)
Taking that equation I did the same but with the x0 and y0 for the points of the center: (x^2+xo)/a^2-(y^2+yo)/b^2=1
I then solved for y: y=b/a(x+xo)+yo
And this shows that beta and alpha are a ratio of each other.
Would this be correct?
Well of course given two angles they have a ratio, that is given two angles a and b, angle a = k*angle b for some real number k. But what is the ratio? Please show us your work.

for the points of the center Are you saying that there are multiple points for the center? If yes, then please list them. Also are you saying that one center is (x0, y0)?

I did: Der(x^2/a^2)-Der(y^2/b^2)=Der(1) . Very good, but can you please tell us what you got in case you made any mistakes?
Taking that equation I did the same but with the x0 and y0 If you replaced x with (x0 and y with y0, then the expression will be a constant, so the derivative will be 0.Not a good thing.

I might be wrong but I think that you may have some mistakes in your work. So can you please show it to us so we can confirm if it valid work. Thanks!

#### MarkFL

##### Super Moderator
Staff member
I believe I was able to solve using calculus.
I did: Der(x^2/a^2)-Der(y^2/b^2)=Der(1)
Taking that equation I did the same but with the x0 and y0 for the points of the center: (x^2+xo)/a^2-(y^2+yo)/b^2=1
I then solved for y: y=b/a(x+xo)+yo
And this shows that beta and alpha are a ratio of each other.
Would this be correct?
I don't think what you did is correct, but your description is too vague to tell. We know the coordinates of the foci are:

$$\displaystyle \left(\pm\sqrt{a^2+b^2},0\right)$$

And so the slope $$m_1$$ of line $$PF_1$$ is:

$$\displaystyle m_1=\frac{y_0}{x_0+\sqrt{a^2+b^2}}$$

And the slope $$m_2$$ of line $$PF_2$$ is:

$$\displaystyle m_2=\frac{y_0}{x_0-\sqrt{a^2+b^2}}$$

Now, the tangent line may be written:

$$\displaystyle y=m\left(x-x_0\right)+y_0$$

Substituting for $$y$$ in the equation of the hyperbola, we obtain:

$$\displaystyle \frac{x^2}{a^2}-\frac{\left(m\left(x-x_0\right)+y_0\right)^2}{b^2}=1$$

We now have a quadratic in $$x$$, which we should now work to arrange in standard form:

$$\displaystyle b^2x^2-a^2\left(m\left(x-x_0\right)+y_0\right)^2=a^2b^2$$

$$\displaystyle b^2x^2-a^2\left(m^2\left(x-x_0\right)^2+2my_0\left(x-x_0\right)+y_0^2\right)=a^2b^2$$

$$\displaystyle b^2x^2-a^2\left(m^2\left(x^2-2x_0x+x_0^2\right)+2my_0\left(x-x_0\right)+y_0^2\right)=a^2b^2$$

$$\displaystyle b^2x^2-a^2\left(m^2x^2-2m^2x_0x+m^2x_0^2+2my_0x-2my_0x_0+y_0^2\right)=a^2b^2$$

$$\displaystyle b^2x^2-a^2m^2x^2+2a^2m^2x_0x-a^2m^2x_0^2-2a^2my_0x+2a^2my_0x_0-a^2y_0^2=a^2b^2$$

$$\displaystyle \left(b^2-a^2m^2\right)x^2+2a^2m\left(mx_0-y_0\right)x-a^2\left(\left(mx_0-y_0\right)^2+b^2\right)=0$$

Now, having arranged the quadratic in standard form, we can equate the discriminant to zero, to find the value of $$m$$, since the tangent line does not touch the hyperbola anywhere else.

$$\displaystyle \left(2a^2m\left(mx_0-y_0\right)\right)^2-4\left(b^2-a^2m^2\right)\left(-a^2\left(\left(mx_0-y_0\right)^2+b^2\right)\right)=0$$

This reduces to:

$$\displaystyle \left(a^2-x_0^2\right)m^2+2x_0y_0m-\left(b^2+y_0^2\right)=0$$

Applying the quadratic formula, we find:

$$\displaystyle m=\frac{-2x_0y_0\pm\sqrt{(2x_0y_0)^2+4\left(a^2-x_0^2\right)\left(b^2+y_0^2\right)}}{2\left(a^2-x_0^2\right)}$$

$$\displaystyle m=\frac{-x_0y_0\pm\sqrt{a^2b^2+a^2y_0^2-b^2x_0^2}}{a^2-x_0^2}$$

Now, since the point $$\left(x_0,y_0\right)$$ is on the hyperbola, we know:

$$\displaystyle \frac{x_0^2}{a^2}-\frac{y_0^2}{b^2}=1$$

And this can be arranged as:

$$\displaystyle a^2b^2+a^2y_0^2-b^2x_0^2=0$$

Hence:

$$\displaystyle m=\frac{x_0y_0}{x_0^2-a^2}$$

Using:

$$\displaystyle a^2b^2+a^2y_0^2-b^2x_0^2=0$$

We find:

$$\displaystyle x_0^2-a^2=\frac{a^2y_0^2}{b^2}$$

And so we may write:

$$\displaystyle m=\frac{x_0y_0}{\dfrac{a^2y_0^2}{b^2}}=\frac{b^2x_0}{a^2y_0}$$

Now, let's find the slope using the calculus:

Implicitly differentiating the hyperbola with respect to $$x$$, we find:

$$\displaystyle \frac{2x}{a^2}-\frac{2y}{b^2}y'=0\implies y'=\frac{b^2x}{a^2y}$$

And so, at the point $$\left(x_0,y_0\right)$$, we have:

$$\displaystyle y'=\frac{b^2x_0}{a^2y_0}\quad\checkmark$$

Okay, we now have the slopes of all three lines, so consider the following diagram:

We know:

$$\displaystyle \arctan\left(m_1\right)+\left(\pi-\arctan(m)\right)+\alpha=\pi\implies \alpha=\arctan(m)-\arctan\left(m_1\right)$$

$$\displaystyle \arctan(m)+\left(\pi-\arctan\left(m_2\right)\right)+\beta=\pi\implies \beta=\arctan\left(m_2\right)-\arctan(m)$$

Next, consider the following identity:

$$\displaystyle \arctan(a)-\arctan(b)=\arctan\left(\frac{a-b}{1+ab}\right)$$

Hence:

$$\displaystyle \alpha=\arctan\left(\frac{m-m_1}{1+mm_1}\right)=\arctan\left(\frac{\dfrac{b^2x_0}{a^2y_0}-\dfrac{y_0}{x_0+\sqrt{a^2+b^2}}}{1+\dfrac{b^2x_0}{a^2y_0}\cdot\dfrac{y_0}{x_0+\sqrt{a^2+b^2}}}\right)$$

$$\displaystyle \beta=\arctan\left(\frac{m_2-m}{1+mm_2}\right)=\arctan\left(\frac{\dfrac{y_0}{x_0-\sqrt{a^2+b^2}}-\dfrac{b^2x_0}{a^2y_0}}{1+\dfrac{b^2x_0}{a^2y_0}\cdot\dfrac{y_0}{x_0-\sqrt{a^2+b^2}}}\right)$$

To simplify, we find:

$$\displaystyle \frac{\dfrac{b^2x_0}{a^2y_0}-\dfrac{y_0}{x_0+\sqrt{a^2+b^2}}}{1+\dfrac{b^2x_0}{a^2y_0}\cdot\dfrac{y_0}{x_0+\sqrt{a^2+b^2}}}=\frac{b^2x_0^2+b^2x_0\sqrt{a^2+b^2}-a^2y_0^2}{a^2x_0y_0+a^2y_0\sqrt{a^2+b^2}+b^2x_0y_0}=\frac{a^2b^2+b^2x_0\sqrt{a^2+b^2}}{\left(a^2+b^2\right)x_0y_0+a^2y_0\sqrt{a^2+b^2}}=\frac{b^2\left(a^2+x_0\sqrt{a^2+b^2}\right)}{y_0\sqrt{a^2+b^2}\left(\sqrt{a^2+b^2}x_0+a^2\right)}=\frac{b^2}{y_0\sqrt{a^2+b^2}}$$

Likewise, we find:

$$\displaystyle \frac{\dfrac{y_0}{x_0-\sqrt{a^2+b^2}}-\dfrac{b^2x_0}{a^2y_0}}{1+\dfrac{b^2x_0}{a^2y_0}\cdot\dfrac{y_0}{x_0-\sqrt{a^2+b^2}}}=\frac{b^2}{y_0\sqrt{a^2+b^2}}$$

And so we may conclude:

$$\displaystyle \alpha=\beta$$

Normally, we do not provide complete solutions, but I found this problem interesting, and once I started, I couldn't stop.

#### Nathan0707

##### New member
Oh wow, I had just figured that when asymptotes are perpendicular a = b and since mine are vertical a = b.. my teacher told us
that we didn’t actually need to know this topic yet at this level and that she didn’t mean to give it to us. For my own curiosity could you show me the non calculus method, if that’s not too much.

#### MarkFL

##### Super Moderator
Staff member
The asymptotes of the given hyperbola are:

$$\displaystyle y=\pm\frac{b}{a}x$$

So, if $$a=b$$, they are indeed perpendicular, but neither would be vertical. We were not given any information regarding the relationship between $$a$$ and $$b$$ though, so they must be treated as general positive real valued parameters.

The pre-calculus method is included in my post above, and is where I let the tangent line be $$\displaystyle y=m\left(x-x_0\right)+y_0$$, substituted that into the equation of the hyperbola, and set the discriminant of the resulting quadratic in $$x$$ to zero because that quadratic could only have one solution given that a tangent line will only touch the hyperbola once. After that, I used the calculus to determine the slope of the tangent line.

#### Nathan0707

##### New member
Oh my teacher meant by non calculus to be proved with like geometry

#### MarkFL

##### Super Moderator
Staff member
What I did is considered analytic or coordinate geometry. The original problem statement stated to solve the problem with and without calculus, which is exactly what I did. I'm satisfied with what I did, but if others can offer alternate solutions, I would be happy to see them.

#### Nathan0707

##### New member
I know you've already done a lot and have really helped me but here is some more clarification my teacher gave: "I intended the non-calculus portion of the question to be a method without the use of calculations. So while the proof indeed belongs to the analytic/coordinate geometry category, it is not what the question intended to test.

For example, You can instead use the triangle inequality, congruent triangles, and some logic to prove that alpha = beta

It is basically proof for the optical property of hyperbola."

#### Dr.Peterson

##### Elite Member
I'm not generally very happy about problems that give only vague indications of what kind of work is expected, but then end up saying, "Do it my way or else." (Math doesn't include mind reading.) Your teacher should at least have said, "Use two different methods, one using calculus, and one using synthetic geometry." But the latter also depends on what geometric definition you are using for the hyperbola. Since the hyperbola was given in terms of an equation, we have to guess at the starting point. I suppose, since both foci are mentioned, most likely the definition based on the difference of the distances to the foci is intended, rather than, say, the focus-directrix definition.

I know a proof for the related property of the parabola that depends on showing that any point on the parabola is on the same side of the putative tangent line, using some facts about triangles. You can see it here. Probably you can do something similar for your problem (for any point on the same branch). Or, more or less equivalently, show that if alpha ≠ beta, then the line intersects the hyperbola.

Perhaps you can also explain what it meant when you said, "my teacher told us that we didn’t actually need to know this topic yet at this level and that she didn’t mean to give it to us." What have you learned in geometry that might be of use? What have you tried?

#### MarkFL

##### Super Moderator
Staff member
Consider $$\triangle F_1PF_2$$. Now, the tangent line is:

$$\displaystyle y=\frac{b^2x_0}{a^2y_0}(x-x_0)+y_0=\frac{b^2x_0}{a^2y_0}x-\frac{b^2x_0^2-a^2y_0^2}{a^2y_0}=\frac{b^2x_0}{a^2y_0}x-\frac{b^2}{y_0}$$

Let $$c=\sqrt{a^2+b^2}$$

Now, the $$x$$-intercept of this line is:

$$\displaystyle \left(\frac{a^2}{x_0},0\right)$$

Let's call this point $$A$$. We then find the length $$d$$ of $$\overline{F_2A}$$ to be:

$$\displaystyle d=c-\frac{a^2}{x_0}$$

And we find the length $$e$$ of $$\overline{F_1A}$$ to be:

$$\displaystyle e=c+\frac{a^2}{x_0}$$

Hence:

$$\displaystyle \frac{d}{e}=\frac{c-\dfrac{a^2}{x_0}}{c+\dfrac{a^2}{x_0}}=\frac{cx_0-a^2}{cx_0+a^2}$$

Consider also the length $$f$$ of $$\overline{PF_2}$$ is:

$$\displaystyle f=\sqrt{\left(x_0-c\right)^2+y_0^2}$$

And let the length $$g$$ of $$\overline{PF_1}$$ is:

$$\displaystyle f=\sqrt{\left(x_0+c\right)^2+y_0^2}$$

Hence:

$$\displaystyle \frac{f}{g}=\frac{\sqrt{\left(x_0-c\right)^2+y_0^2}}{\sqrt{\left(x_0+c\right)^2+y_0^2}}$$

If we observe that:

$$\displaystyle y_0^2=\frac{b^2}{a^2}x_0^2-b^2$$

Then we have:

$$\displaystyle \left(x_0-c\right)^2+y_0^2=\left(x_0-c\right)^2+\frac{b^2}{a^2}x_0^2-b^2=x_0^2-2cx_0+c^2+\frac{b^2}{a^2}x_0^2-b^2=x_0^2-2cx_0+a^2+\frac{b^2}{a^2}x_0^2=\frac{c^2}{a^2}x_0^2-2cx_0+a^2$$

Likewise:

$$\displaystyle \left(x_0+c\right)^2+y_0^2=\frac{c^2}{a^2}x_0^2+2cx_0+a^2$$

And so:

$$\displaystyle \frac{f}{g}=\sqrt{\frac{\frac{c^2}{a^2}x_0^2-2cx_0+a^2}{\frac{c^2}{a^2}x_0^2+2cx_0+a^2}}=\sqrt{\frac{c^2x_0^2-2a^2cx_0+a^4}{c^2x_0^2-2a^2cx_0+a^4}}=\sqrt{\left(\frac{cx_0-a^2}{cx_0+a^2}\right)^2}=\frac{cx_0-a^2}{cx_0+a^2}=\frac{d}{e}$$

And so, by the angle bisector theorem, we may conclude that:

$$\displaystyle \alpha=\beta$$