1. ## Hyperbola

Find the acute angle between the tangents which can be drawn from (2,3) to the hyperbolas x^2 - y^2 = 2. I've found dy/dx = x/y but I don't know what to do with it. Do I simply sub in (2,3)? If so, what do I do next???

2. Originally Posted by xwrathbringerx
Find the acute angle between the tangents which can be drawn from (2,3) to the hyperbolas x^2 - y^2 = 2. I've found dy/dx = x/y but I don't know what to do with it. Do I simply sub in (2,3)? If so, what do I do next???
Well, you certainly don't substitute x= 2, y= 3 into the equation of the hyperbola because (2, 3) is NOT on the hyperbola! At any point, $\displaystyle (x_0, y_0)$, the slope of the tangent line is given by $\displaystyle m= \frac{dy}{dx}= \frac{x_0}{y_0}$ so the equation of a line that is tangent to the hyperbola at $\displaystyle (x_0, y_0)$ is $\displaystyle y= \frac{x_0}{y_0}(x- x_0)+ y_0$.

Now, since that tangent line must pass through (2, 3), you can replace x by 2 and y by 3 in that: $\displaystyle 3= \frac{x_0}{y_0}(2- x_0)+ y_0$. Multiplying through by $\displaystyle y_0$ to get rid of the fraction, $\displaystyle 3y_0= 2x_0- x_0^2+ y_0^2$ or $\displaystyle x_0^2- y_0^2= 2x_0- 3y_0$, a quadratic equation in $\displaystyle x_0$ and $\displaystyle y_0$. Since $\displaystyle (x_0, y_0)$ is on the hyperbola, you also have $\displaystyle x_0^2- y_0^2= 2$ so that $\displaystyle x_0^2- y_0^2= 2x_0- 3y_0= 2$ and then $\displaystyle y_0= (2/3)x_0- 2/3$. You can substitute that into either of the previous equations to get a quadratic equation for $\displaystyle x_0$ which, hopefully, will have two roots giving the two points $\displaystyle (x_0, y_0)$ where the two tangents through (2, 3) touch the hyperbola. That, of course, will give you the slopes, $\displaystyle \frac{x_0}{y_0}$, of the two lines and you can get the angle between them from that.

3. Hmmmmm the coordinates I get are basically quadratic formula forms...how on earth do I solve that to get the angle??? is there any other possible "easier" way?