The hyperbola $\displaystyle \frac{x^2}{a^2}-\frac{y^2}{b^2} = 1 $ has an eccentricity of e. The ellipse $\displaystyle \frac{x^2}{a^2+b^2}+\frac{y^2}{b^2} = 1$ has an eccentricity of 1/e.
If the two graphs intersect at P in the first quadrant, show that the acute angle $\displaystyle \Theta $ between the the tangents to the curves at P satisfies:
$\displaystyle tan\Theta = \sqrt{2}(e + \frac{1}{e}) $
Thanks!
(Angry)
Use implicit differentiation to find the slope of the tangent line of each graph at point (x,y). The slope is, of course, the tangent of the angle each makes with the x-axis. And the angle between the tangent lines (which is, by definition, the angle between the curves) is the difference between those two angles. Use the trig indentity $\displaystyle tan(x- y)= \frac{tan(x)- tan(y)}{1+ tan(x)tan(y)}$.
You will need to know, if you do not already, that the eccentricity of the ellipse, $\displaystyle \frac{x^2}{a^2}+ \frac{y^2}{b^2}= 1$, is given by $\displaystyle e= \frac{\sqrt{a^2- b^2}}{a}$ (for a> b) and that the eccentricity of the hyperbola, $\displaystyle \frac{x^2}{a^2}- \frac{y^2}{b^2}= 1$, is given by $\displaystyle e= \frac{\sqrt{a^2+ b^2}}{a}$ (also for a> b).
for your eccentricities: shouldn't those be divided by a?Quote:
Originally Posted by HallsofIvy
Yes, thanks. I have gone back and corrected my post. What I gave was the "linear eccentricity".Quote:
Originally Posted by Jhevon
