Hyperbola Focus Length Greater than Semi-Major Axis: Is it a Necessity?

[bananabandana]

Homework Statement

Why is it necessarily true that for a hyperbola, the focus length, $f$ has got to be greater than the semi-major axis , $a$ - $f >a$ ?

Homework Equations

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The Attempt at a Solution

I needed to derive the cartesian equation of a hyperbola with centre at $(\alpha,\beta)$ and foci along the $\alpha -$ axis. using the definition that the difference between the distances to the two foci is a constant. After some algebra, I just got back the equation of an ellipse.

$$ \frac{(x-\alpha)^{2}}{a^{2}}+\frac{(y-\beta)^{2}}{a^{2}-f^{2}}=1 $$

Of course I realize that this is also the equation of a hyperbola, under the condition that $f>a$ so that :
$$ \frac{(x-\alpha)^{2}}{a^{2}}-\frac{(y-\beta)^{2}}{a^{2}-f^{2}}=1 $$

But I'm stuck at proving that this condition is, in general true, just from the definition given.

Could someone give me a hint?

Thanks!

 

[mfb]

I think you forgot to swap the signs in the denominator in the second equation.

We just call shapes with f>a "hyperbola" and with f<a "ellipse", they get different names because they look so different. In particular, a hyperbola has solutions for arbitrarily large x and y, which means the two summands must have different signs, whereas they have the same sign for an ellipse.

 

[bananabandana]

Yes, sorry I did! So the definition of the hyperbola as being the locus where there is a constant difference between the distances to two foci is incomplete? I need to know that $f>a$ on top - I can't prove that just from this definition?

 

[mfb]

f>a should follow from that requirement.

 

[bananabandana]

That's my difficulty, I don't know how to show that it must follow... :P

In full my working currently looks like this - I'll borrow this convenient wolfram diagram for the symbols. Although the diagrams show clearly that $f>a$ I do not think I assume this in my calculation. I've dropped the shift in the centre of the co-ordinate system because it doesn't really matter and obscures the working. (so centre is now (0,0) not $(\alpha, \beta )$

Let $r_{2}= F_{2}P \ , \ r_{1} = F_{1}P$

$$ r_{2}-r_{1} =C \implies \sqrt{(x-f)^{2}+y^{2}}-\sqrt{(x+f)^{2}+y^{2}} =C $$
$$ \therefore (x-f)^{2}+y^{2} = C^{2}+(x+f)^{2}+y^{2}+2C\sqrt{(x+f)^{2}+y^{2}} $$
$$ \implies -4xf=C^{2}+2C \sqrt{(x+f)^{2}+y^{2}} $$
$$ (C+4x \frac{f}{C})^{2}=4(x+f)^{2}+4y^{2} $$
$$ C^{2}+16\frac{x^{2}f^{2}}{C^{2}}+8xf = 4x^{2}+8xf+4f^{2} + 4y^{2} $$
$$ C^{2}-4f^{2} = 4x^{2}(\frac{C^{2}-4f^{2}}{C^{2}})+4y^{2} \implies \frac{4x^{2}}{C^{2}}+\frac{4y^{2}}{C^{2}-4f^{2}} =1 $$

From above ,at $y =0 \ , \ 2x =C$ By definition of the semi-major axis of the hyperbola, if $y= 0 \implies x=a$

$$ \therefore \frac{x^{2}}{a^{2}}+\frac{y^{2}}{a^{2}-f^{2}}=1 $$

But from here I can't see how to retrieve the minus sign.
Thanks!

 

[mfb]

If C>2f (which corresponds to f<a as 2a=C), you do not get any solution of your original equation. The squaring process fakes an additional solution which has a meaning for an ellipse, but not here.