determining fixed angle

[synx]

my question is:
Determine the fixed angle between the two tangent lines to x^2-y^2=3 and x^2-4x+y^2=0 at their point of intersection to the nearest degree.

ok, so the derivative of each of those is:
y=x , and y=-x+2

So how do I the degree of intersection between those 2?

 

[NYC]

my question is:
Determine the fixed angle between the two tangent lines to x^2-y^2=3 and x^2-4x+y^2=0 at their point of intersection to the nearest degree.

ok, so the derivative of each of those is:
y=x , and y=-x+2

So how do I the degree of intersection between those 2?

Since you're dealing with x's and y's in this manner, this is an implicit derivative, and thus: dy/dx=x/y (where you said x=y) and dy/dx in the second case= (2-x)/y

OK, this should give you the slopes of the tangent lines, assuming you can find x and y...which you can do by taking a look ar your original equations. Essentially, you need that point of intersection (x,y) , so you can evaluate the slopes of the tangent lines. y squared appears in both, so first establish that y squared equals x squared - 3 (by rearranging the first equation), then substitute this into x squared - 4x - y squared = 0 ; you will end up with 2x^2-4x-3=0. Since this does not factor, use the quadratic equation to find x, then plug those results back into the equation to get y. ONCE YOU HAVE X AND Y, go to those derivatives (dy/dx's) and plug in, and you should come up with the slopes. In terms of finding the angle, take the inverse tangent of each slope, then subtract the angles which are generated. (Slope is change in y over change in x, and if you look at the graph, tangent of an angle is the same thing) Your lines will of course, have a slope, and thus an angle of inclination. Anyway, this is the approach I would take. I hope this was of some help.

 

[skeeter]

you sure you have the original equations for the hyperbola and the circle correct?

reason I ask, is that the two points of intersection are not nice ...

x = 1 + [sqrt(10)/2], and y = +/- sqrt[(2sqrt(10) + 1)/2]

 

[soroban]

Hello, synx!

Determine the fixed angle between the two tangent lines to \(\displaystyle x^2\,-\,y^2\:=\:3\) and \(\displaystyle x^2\,-\,4x\,+\,y^2\:=\:0\)
at their point of intersection to the nearest degree.

ok, so the derivative of each of those is:
y = x , and y = -x + 2 \(\displaystyle \;\;\) . . . ?? . . . how did you get these?

First, find their intersections (there are two).
\(\displaystyle \;\;\)Solve the system of equationsL \(\displaystyle \:\begin{array}{cc}x^2\,-\,y^2\:=\:3 \\ x^2-\,4x\,+\,y^2\:=\:0\end{array}\)

Then find their derivatives so we can determine slopes at the intersection(s).
\(\displaystyle \;\;2x\,-\,2y\left(\frac{dy}{dx}\right)\:=\:0\;\;\Rightarrow\;\;\frac{dy}{dx}\:=\:\frac{x}{y}\)
\(\displaystyle \;\;2x\,-\,4\,+\,2y\left(\frac{dy}{dx}\right)\:=\:0\;\;\Rightarrow\;\;\frac{dy}{dx}\:=\:\frac{2\,-\,x}{y}\)


Once you have the two slopes, \(\displaystyle m_1\) and \(\displaystyle m_2\), find the angle \(\displaystyle \theta\) between them with:

\(\displaystyle \L\;\;\tan\theta\;=\;\frac{m_2\,-\,m_1}{1\,+\,m_1m_2}\)

 

[synx]

yeah, sorry skeeter the correct equations are x^2-y^2=3 and x^2-4x+y^2+3=0

Ok, i think i understand how to do it, but at the same time i think im stuck.

y^2=x^2-3 , x^2-4x+y^2+3=0

I plug in y^2 to the second equation which simplifys down to 2x(x-2)=0 and get the roots 2 and 0.

So i plug these back into the first equations and only root 2 works in each, 0 just gives the square root of a negative.

eq#1 f(2) = 1 , so (2,1)
deriv#1 = x/y, (2/1) = 2
eq#2 f(2) = 1, also (2,1)
deriv#2 = (2-x)/y , plugging in 2 gives zero. Did i do something wrong?