1. ## Re: Proof involving hyberbola

Yes you ar on the right track but to get from line 1 to line 2 you were multiplying throughout by a^2b^2. So instead of =1 you should have =a^2b^2
ax^2+bx+c=0 has repeated roots if b^2-4ac=0 so apply this to your corrected equation.

2. ## Re: Proof involving hyberbola

You are multiplying both sides of the equation by a^2b^2 so in your second line you should have =a^2b^2 instead of =1

3. ## Re: Proof involving hyberbola

OK, here goes:

a = (b2 - m2a2)
b = 2mca2
c = (a2b2 - c2)

For repeated roots, b2 = 4ac

4m2c2a2 = 4.(b2 - m2a2)(a2b2 - c2)

If I try to simplify and solve this equation, I don't get the result b2 - 4ac = 0

If I simply ignore the constants a,b & c, I get the equation x2+2x+1 = 0, for which I get the roots (x+1) & (x+1)

Is this correct?

4. ## Re: Proof involving hyberbola

yes right track but your second line should be=a^2b^2 (You were multiplying both sides by a^2b^2)
Now the general quadratic ax^2+bx+c=0 has repeated roots if b^2-4ac=0

5. ## Re: Proof involving hyberbola

So,

$\displaystyle x^2(b^2 - m^2a^2) + x(-2mca^2) + (-a^2c^2- a^2b^2) = 0$

a = (b2 - m2a2)
b = -2mca2
c = (- a2c2 - a2b2)

b2 = 4ac

4m2c2a4 = 4.(b2 - m2a2).(-a2c2 - a2b2)

4m2c2a4 = 4[ -a2b2c2 + m2a4c2 - a2b4 + m2a4b2]

The two sides are not equal.

We get 4m2c2a4 on both sides, but what to do with the other terms?

6. ## Re: Proof involving hyberbola

That 1 in your second line should be a^2b^2 (You were in the process of multiplying both sides by a^2b^2
Now ax^2+bx+c=0 has repeated roots if b^2-4ac=0
So in our equation want (2a^2mc)^2+4(b^2-a^2m^2)a^2(b^2+c^2)=0

7. ## Re: Proof involving hyberbola

That 1 in your second line should be $\displaystyle a^2b^2$ (You were in the process of multiplying both sides by $\displaystyle a^2b^2$
So line should read $\displaystyle (b^2-a^2m^2)x^2-2a^2mcx-a^2(b^2+c^2)=0$
Now $\displaystyle ax^2+bx+c=0$ has repeated roots if $\displaystyle b^2-4ac=0$
So in our equation want $\displaystyle (2a^2mc)^2+4(b^2-a^2m^2)a^2(b^2+c^2)=0$

I have carried out the correction of multiplying RHS also by $\displaystyle a^2b^2$ and started with the corrected equation,

$\displaystyle x^2(b^2 - m^2a^2) + x(-2mca^2) + (-a^2c^2- a^2b^2) = 0$

which can be rewritten as

$\displaystyle x^2(b^2 - m^2a^2) + x(-2mca^2) -a^2(c^2 + b^2) = 0$

I have carried out the multiplication as you have said. Incidentally, the equation should be

$\displaystyle (2a^2mc)^2-4(b^2-a^2m^2)a^2(b^2+c^2)=0$

and not

$\displaystyle (2a^2mc)^2+4(b^2-a^2m^2)a^2(b^2+c^2)=0$

because the term "C" is negative.

which I have correctly done. How does all this relate to the original equation $\displaystyle p^2 = a^2cos^2 A - b^2 sin ^2 A$?

Edit: Whether the term is negative or positive, it does not matter since there are uncancelled terms on the RHS.

8. ## Re: Proof involving hyberbola

This is correct. Now multiply out brackets. There will be an a^2 which can be cancelled out of every term (since right hand side=0)
You should also see among the terms a^2m^2c^2-a^2m^2c^2, so these go.
Now you should see that there is b^2 which can be cancelled.
Should now have b^2+c^2-a^2m^2=0
Now replace c and m by what we let them stand for at the beginning.

9. ## Re: Proof involving hyberbola

OK, I got it, thanks again for your patience.

10. ## Re: Proof involving hyberbola

Well done! Incidentally in your last post you said that the 'C' is negative which is correct but the formula for repeated roots is B^2-4AC so we had two minus signs which is why I finished with a plus sign.

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