# Math Help - Hyperbola/Tangent proof

1. ## Hyperbola/Tangent proof

I have been wrestling with this for about 45 mins now, I don't think it's that hard but I can't quite get it.

Prove that the line with equation $y=mx+c$ is a tangent to the hyperbola with equation $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ if, and only if $a^2m^2=b^2+c^2$

Simple enough, right? But I ended up with two A4 sides of working (which I won't repeat here) and a complete mess. My approach was to prove that if it was a tangent then when I solved the two equations simultaneously the discriminant of the resulting quadratic would be equal to zero (since the tangent only touches once), I was hoping to rearrange this to get the required relationship between a,m,b and c. This didn't work as expected.

I also have no idea how to prove the converse

Any help would be great! Thanks!

2. Substituting:
x^2 / a^2 - (mx + c)^2 / b^2 = 1

Simplifying (if you can call it that!):
x^2(a^2 m^2 - b^2) + 2 a^2 cmx + a^2(b^2 + c^2) = 0

Substituting a^2 m^2 = b^2 + c^2 in 1st term, then b^2 + c^2 = a^2 m^2 in 3rd term:
x^2 c^2 + 2 a^2 cmx + a^4 m^2 = 0

x = [-2 a^2 cm +- SQRT(4 a^4 c^2 m^2 - 4 a^4 c^2 m^2)] / (2 c^2)
There ya go: zero discriminant!

x = -a^2 m / c

I'm not saying I understand all that's going on... but I am proud to have
arrived at a zero discriminant

3. Originally Posted by Stonehambey
I have been wrestling with this for about 45 mins now, I don't think it's that hard but I can't quite get it.

Prove that the line with equation $y=mx+c$ is a tangent to the hyperbola with equation $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ if, and only if $a^2m^2=b^2+c^2$

Simple enough, right? But I ended up with two A4 sides of working (which I won't repeat here) and a complete mess. My approach was to prove that if it was a tangent then when I solved the two equations simultaneously the discriminant of the resulting quadratic would be equal to zero (since the tangent only touches once), I was hoping to rearrange this to get the required relationship between a,m,b and c. This didn't work as expected.

I also have no idea how to prove the converse

Any help would be great! Thanks!

Your approach looks OK to me. Post the highlights of your working.

4. Originally Posted by mr fantastic
Your approach looks OK to me. Post the highlights of your working.
Sure,

$y=mx+c$ and $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

$\frac{x^2}{a^2}-\frac{(mx+c)^2}{b^2}=1$

$\frac{x^2b^2-a^2(m^2x^2+2mcx+c^2)}{a^2b^2}=1$
.
.
.
$(b^2-a^2m^2)x^2-(2a^2mc)x-(c^2+a^2b^2)=0$

We require $B^2-4AC=0$, that is,

$4a^4m^2c^2+4(b^2-a^2m^2)(c^2+a^2b^2)=0$
.
.
.
$a^2(m^2c^2a^2+b^4)+b^2c^2=a^2m^2(c^2+a^2b^2)$

I messed around with the above equation for a while before concluding that the mistake had already been made

5. Originally Posted by Stonehambey
Sure,

$y=mx+c$ and $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

$\frac{x^2}{a^2}-\frac{(mx+c)^2}{b^2}=1$

$\frac{x^2b^2-a^2(m^2x^2+2mcx+c^2)}{a^2b^2}=1$
.
.
.
$(b^2-a^2m^2)x^2-(2a^2mc)x-(c^2+a^2b^2)=0$ Mr F says: This line is wrong. It should be $(b^2-a^2m^2)x^2-(2a^2mc)x-({\color{red}a^2} c^2+a^2b^2)=0$.

[snip]
Now get the discriminant and equate it to zero. Then expand and simplify.

6. Ah, thanks, that worked.

So that shows "if tangent then relationship", does Wilmer's reply show "if relationship then tangent", since he subbed in the relationship to the equation and show that the discriminant was zero (so a tangent)?

7. Originally Posted by mr fantastic
Now get the discriminant and equate it to zero. Then expand and simplify.
MUCH easier to do the substituting first...as I did...

8. Originally Posted by Stonehambey
So that shows "if tangent then relationship", does Wilmer's reply show "if relationship then tangent", since he subbed in the relationship to the equation and show that the discriminant was zero (so a tangent)?