Proof that any tangent to an hyperbola bisects the angle of the point with the foci

Question

I'm trying to solve this problem in my text book.

The approach I want to try is as follows:

1. The tangent vector at P is $\vec{v_h} = (1, y')$. Where y' is the derivative of the hyperbola function.
2. Given $\vec{u_1} = \frac{\vec{F'P}} {||\vec{F'P}||}$ and $\vec{u_2} = \frac{\vec{FP}} {||\vec{FP}||}$, these dot products should be equal $\vec{v_h} \cdot \vec{u_1} = \vec{v_h} \cdot \vec{u_2} $ because they are supposed to form the same angle with the tangent line.

My attempt:

Start with the equation of the hyperbola: $$ b^2x^2 - a^2y^2=a^2b^2 $$

In this formula, c is half the distance between the foci.

Then isolate y:

$$ y = \pm \sqrt{\frac{b^2 x^2 - a^2 b^2}{a^2}} $$

Then we compute the derivative of y which tells us the slope of the function.

$$ y' = \frac{b^2}{a^2} \frac{x}{y} $$

With the derivative we can get a direction vector for the slope. $$ \vec{v_h} = (1, y') $$

And we compute the unit vectors that go from the foci to P.

$$ \vec{u_1} = \frac{1}{\sqrt{(c+x)^2 + y^2}} (c+x, y) $$ $$ \vec{u_2} = \frac{1}{\sqrt{(c-x)^2 + y^2}} (c-x, y) $$

If the angles between he tangent and the foci are the same, the cosine of the angle is too, and in this case, the dot product is the cosine of the angle scaled by the length of $\vec{v_h}$.

$$ \vec{v_h} \cdot \vec{u_1} = \vec{v_h} \cdot \vec{u_2} $$

$$ \frac{c+x + y y'} {\sqrt{(c+x)^2 + y^2}} = \frac{c-x + y y'}{\sqrt{(c-x)^2 + y^2}} $$

From this point, I have tried many things but I'm unable to prove that both sides are equal. Can you please help me?

Answer 1

It took me a while but managed to solve it.

This equation was wrong.

$$ \frac{c+x + y y'} {\sqrt{(c+x)^2 + y^2}} = \frac{c-x + y y'}{\sqrt{(c-x)^2 + y^2}} $$

It should be

$$ \frac{c+x + y y'} {\sqrt{(c+x)^2 + y^2}} = \frac{x-c + y y'}{\sqrt{(x-c)^2 + y^2}} $$

We know that

$$ y = \pm \sqrt{\frac{b^2 x^2 - a^2 b^2}{a^2}} $$

$$ y' = \frac{b^2}{a^2} \frac{x}{y} $$

We are going to substitute the $b$ using this equation

$$ b^2 = c^2 - a^2 $$

Then

$$ y = \pm \sqrt{\frac{c^2 - a^2}{a^2}x^2 - (c^2 - a^2)} $$

$$ y' = \frac{c^2-a^2}{a^2} \frac{x}{y} $$

If we substitute $y'$ in the first equation, $y$ cancels out nicely.

$$ \frac{c+x + \frac{c^2-a^2}{a^2}x} {\sqrt{(c+x)^2 + y^2}} = \frac{x-c + \frac{c^2-a^2}{a^2}x}{\sqrt{(x-c)^2 + y^2}} $$

Rearange fractions $$ \frac{\sqrt{(x-c)^2 + y^2}}{\sqrt{(x+c)^2 + y^2}}= \frac{\frac{c^2}{a^2}x-c}{\frac{c^2}{a^2}x+c} $$

Square both sides

$$ \frac{(x-c)^2 + y^2}{(x+c)^2 + y^2}= \left[\frac{\frac{c^2}{a^2}x-c}{\frac{c^2}{a^2}x+c} \right] ^2 $$

Expand the squares

$$ \frac{x^2 + c^2 - 2cx + y^2}{x^2 + c^2 + 2cx + y^2} = \frac{\frac{c^4}{a^4}x^2 + c^2 -2\frac{c^3}{a^2}x}{\frac{c^4}{a^4} x^2 + c^2 +2\frac{c^3}{a^2}x} $$

On the right hand side, divide numerator and denominator by $c^2$

$$ \frac{x^2 + c^2 - 2cx + y^2}{x^2 + c^2 + 2cx + y^2} = \frac{\frac{c^2}{a^4}x^2 + 1 -\frac{2c}{a^2}x}{\frac{c^2}{a^4} x^2 + 1 +\frac{2c}{a^2}x} $$

Move denominators to the other side $$ (x^2 + c^2 - 2cx + y^2) (\frac{c^2}{a^4} x^2 + 1 +\frac{2c}{a^2}x) = \\ (x^2 + c^2 + 2cx + y^2) (\frac{c^2}{a^4}x^2 + 1 -\frac{2c}{a^2}x) $$

Now we do the multiplications. Most terms are equal on both sides so they cancel out.

$$ \frac{2c}{a^2} x^3 + \frac{2c^3}{a^2} x - \frac{2c^3}{a^4} x^3 - 2cx + \frac{2c}{a^2}xy^2 = \\ -\frac{2c}{a^2} x^3 - \frac{2c^3}{a^2} x + \frac{2c^3}{a^4} x^3 + 2cx - \frac{2c}{a^2}xy^2 $$

Move everything to the same side and divide by $4cx$

$$ \frac{x^2}{a^2} + \frac{c^2}{a^2} - \frac{c^2}{a^4}x^2 - 1 + \frac{y^2}{a^2} = 0 $$

Gather fractions $$ \frac{x^2 + c^2 + y^2 - a^2}{a^2} - \frac{c^2}{a^4}x^2 = 0 $$

Multiply by $a^2$, then substitute $y^2$

$$ x^2 + c^2 +\frac{c^2 - a^2}{a^2}x^2 - c^2 +a^2 -a^2 -\frac{c^2}{a^2}x^2 = 0 $$

Now we see a few terms cancel out.

$$ x^2 +\frac{c^2 - a^2}{a^2}x^2 -\frac{c^2}{a^2}x^2 = 0 $$

Take $x^2$ as common factor

$$ \frac{c^2 - a^2}{a^2}x^2 -(\frac{c^2}{a^2}-1)x^2 = 0 $$

Substract the 1 to the fraction

$$ \frac{c^2 - a^2}{a^2}x^2 -\frac{c^2 - a^2}{a^2}x^2 = 0 $$

Finally

$0 = 0$

So the angles are the same.