Talk:HSC Extension 1 and 2 Mathematics/4-Unit/Conics

Hyperbola $$H$$ has equation $$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$$ and eccentricity $$e$$. Ellipse $$E$$ has equation $$\frac{x^2}{(a^2+b^2)} + \frac{y^2}{b^2} = 1$$.

i. Show that ellipse $$E$$ has eccentricity $$\frac{1}{e}$$ [1 mark] ii. If $$H$$ and $$E$$ intersect at $$P$$ in the first quadrant, show that the acute angle $$\alpha$$ between the tangents to $$H$$ and $$E$$ at $$P$$ is given by $$\tan \alpha = \sqrt{2}(e + \tfrac{1}{e})$$ [6 marks]

For part ii., here's my working:
 * 1) $$P(a\sec\theta,b\tan\theta)\,$$, from the definition of a hyperbola
 * 2) $$P(\sqrt{a^2+b^2}\cos\theta,b\sin\theta)$$, from the definition of an ellipse ($$a$$ in the ellipse = $$\sqrt{a^2+b^2}$$)
 * 3) $$\therefore b\tan\theta = b\sin\theta$$
 * 4) $$\tan\theta = \sin\theta\,$$
 * 5) $$\cos\theta = \pm 1$$
 * 6) $$\sqrt{a^2+b^2}cos\theta = a\sec\theta$$
 * 7) $$\sqrt{a^2+b^2} = a$$ (since $$\cos\theta = \pm 1$$)
 * 8) $$a^2+b^2 = a^2$$
 * 9) $$b^2 = 0$$