#### Nathan0707

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- Thread starter Nathan0707
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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?

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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.

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 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!

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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:

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?

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

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.

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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.

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\(\displaystyle y=\pm\frac{b}{a}x\)

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.

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

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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."

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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

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

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:

\(\displaystyle \alpha=\beta\)