1. ## Proof involving hyberbola

Prove that the line $x cos \alpha + y sin \alpha = p$ touches the hyperbola $x^2/a^2 - y^2/b^2 = 1$ if $p^{2} = a^2 cos^2 \alpha - b^2 sin^2 \alpha$.

How do I start?

Edit: Sorry for the error in the question.

2. ## Re: Proof involving hyberbola

Check the wording of the question. What you've written isn't a line.

3. ## Re: Proof involving hyberbola

Corrected the question.

4. ## Re: Proof involving hyberbola

Again have you copied the question correctly. That is an ellipse not a hyperbola.

5. ## Re: Proof involving hyberbola

biffboy, thanks for your patience. Error again and corrected again.

6. ## Re: Proof involving hyberbola

Make x the subject of the line equation and substitute this into the hyperbola equation. Rearrange this as a quadratic in y. For the line to be a tangent we require this quadratic to have repeated roots. Applying this condition should produce the result stated.

7. ## Re: Proof involving hyberbola

So, do I substitute the value of y in the line equation with the value of y with respect to x?

In that case, the equation becomes,

So the line equation will become

I tried to simplify the equation and it is becoming messy. Am I doing things right?

(Sorry for the small image, laTEX is pain and I do not know how to put the special characters here. Tried to use an online latex editor but there seems to be some problem)

8. ## Re: Proof involving hyberbola

Replace y in the hyperbola equation with (p-xcosa)/sina

9. ## Re: Proof involving hyberbola

$\displaystyle (p - ysin\alpha/cos\alpha )^2/a^2 - y^2/b^2 = 1$

$\displaystyle (p^2 + y^2sin^2\alpha - 2pysin\alpha)/a^2 cos^2 \alpha - y^2/b^2 = 1$

$\displaystyle (p^2 + y^2sin^2\alpha - 2pysin\alpha)/a^2 cos^2 \alpha = 1 + y^2/b^2$

$\displaystyle (p^2 + y^2sin^2\alpha - 2pysin\alpha)b^2 = (b^2 + y^2)(a^2 cos^2 \alpha)$

This doesn't seem to be leading anywhere. Is this what I am supposed to do?

10. ## Re: Proof involving hyberbola

Rearrange your last line to get a quadratic in y. You then want this quadratic to have a repeated root.

11. ## Re: Proof involving hyberbola

What you have substituted for x should be p/cosa-(sina/cosa)y

12. ## Re: Proof involving hyberbola

So I got the equation

$\displaystyle \left (p - y sin\alpha \right )^2 = (a^2+b^2)(a^2cos^2\alpha )/b^2$

Is this correct?

13. ## Re: Proof involving hyberbola

I can't see what's happened there. Looking back your last line of post 9 was correct so look at my post 10 again

14. ## Re: Proof involving hyberbola

If you want one last fresh start try this. I'll write A instead of alpha. Line can be written y=-(cosA/sinA)x+p/sinA
This is y=mx+c with m=-cosA/sinA and c=p/sinA
Substitute y=mx+c into the hyperbola. So x^2/a^2-(mx+c)^2/b^2=1
Multiply throughout by a^2b^2 and remove the brackets and then rearrange as a quadratic in x
Apply the condition for this to have repeated roots
After some symplifying this gives b^2+c^2-(a^2m^2)=0
Put back in what m and c stood for to get the required result

15. ## Re: Proof involving hyberbola

OK, here's my attempt:

$\displaystyle x^2/a^2 - (mx + c)^2/b^2 = 1$

Multiplying both the terms by a2b2,

$\displaystyle x^2b^2 - (m^2x^2 + c^2 + 2mxc)a^2 = 1$

Rearranging, I get

$\displaystyle x^2(b^2 - m^2a^2) + x(-2mca^2) + (-a^2c^2-1) = 0$

This is of the form ax2+bx+c=0

What do I do now? I don't know how to make it have repeated roots.

I hope I am on the right track.

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