Cartesian form of a Hyperbola

In summary: If C<2f, then there is a real solution and it is (x,y) where x>a and y>-a.Yes, sorry I got confused about the signs.When C=2f, then the equation becomes (x,y) where x<a and y<-a.
  • #1

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!
 
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  • #2
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.
 
  • #3
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?
 
  • #4
f>a should follow from that requirement.
 
  • #5
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 ) ##

HyperbolaFoci_750.gif

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!
 
  • #6
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.
 

1. What is the Cartesian form of a hyperbola?

The Cartesian form of a hyperbola is a type of equation that describes the shape of a hyperbola on a Cartesian plane. It is written in the form (x-h)^2/a^2 - (y-k)^2/b^2 = 1, where (h,k) represents the center of the hyperbola and a and b are the distances from the center to the vertices along the x and y axes, respectively.

2. How is the Cartesian form of a hyperbola different from the general form?

The general form of a hyperbola is written in the form Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0, where A, B, C, D, E, and F are constants. The Cartesian form is a simplified version of the general form and is easier to graph and understand.

3. What does the sign of the constants a and b in the Cartesian form represent?

The sign of a determines the direction of the branches of the hyperbola along the x axis, while the sign of b determines the direction of the branches along the y axis. If a is positive, the branches open to the left and right, and if a is negative, they open up and down. Similarly, if b is positive, the branches open up and down, and if b is negative, they open to the left and right.

4. How can the eccentricity of a hyperbola be determined from its Cartesian form?

The eccentricity of a hyperbola is equal to the square root of a^2 + b^2. This can be found from the distance between the center and the vertices, as a and b are the distances from the center to the vertices along the x and y axes, respectively.

5. Can the Cartesian form of a hyperbola have a center at the origin (0,0)?

Yes, the center of a hyperbola can be at the origin if the values of h and k in the Cartesian form are both equal to 0. In this case, the equation reduces to x^2/a^2 - y^2/b^2 = 1, and the branches of the hyperbola open in opposite directions along the x and y axes.

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