Prove that the line touches the hyperbola if .
How do I start?
Edit: Sorry for the error in the question.
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.
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)
$\displaystyle (p - ysin\alpha/cos\alpha )^2/a^2 - y^2/b^2 = 1$
$\displaystyle (p^2 + y^2sin^2\alpha - 2pysin\alpha)/a^2 cos^2 \alpha - y^2/b^2 = 1$
$\displaystyle (p^2 + y^2sin^2\alpha - 2pysin\alpha)/a^2 cos^2 \alpha = 1 + y^2/b^2$
$\displaystyle (p^2 + y^2sin^2\alpha - 2pysin\alpha)b^2 = (b^2 + y^2)(a^2 cos^2 \alpha)$
This doesn't seem to be leading anywhere. Is this what I am supposed to do?
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
OK, here's my attempt:
$\displaystyle x^2/a^2 - (mx + c)^2/b^2 = 1$
Multiplying both the terms by a^{2}b^{2},
$\displaystyle x^2b^2 - (m^2x^2 + c^2 + 2mxc)a^2 = 1$
Rearranging, I get
$\displaystyle x^2(b^2 - m^2a^2) + x(-2mca^2) + (-a^2c^2-1) = 0$
This is of the form ax^{2}+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.