Conic Sections

2.a) Sketch the ellipse $x^2 + \frac{y^2}{4} = 1$ as shown in Fig 7.6

b) Sketch the ellipse obtained by a horizontal translation of +3 units.

c) Write down the equation of the new ellipse.

d) Find the co-ordinates of the points of intersection of the line y = 2x - 4 and the new ellipse.

I'm stuck on d, its x = 2,3 (yesterday one c was incorrect)

You used y=2x + 4 instead of y=2x - 4

3. Find the points of intersection of the hyperbola x² - y² = 1 and the line y = ⅓(x + 2)

Answer is x =1.55, -1.05, yet I get below a 0 on $b^2 - 4ac$
What did I screw up this time?

3. Find the points of intersection of the hyperbola x² - y² = 1 and the line y = ⅓(x + 2)

Answer is x =1.55, -1.05, yet I get below a 0 on $b^2 - 4ac$
What did I screw up this time?

Well you screwed up on calculating the determinant. The quadratic is correct.

i think that the discriminant would be more use in this problem...

Too much linear algebra lately

Well you screwed up on calculating the discriminant. The quadratic is correct.

The quadratic is correct

...

You forgot that $\arccos(\frac{\sqrt{3}}{2}) =2\pi n \pm \frac{\pi}{6}, \forall n \in \mathbb{Z}$

$[br] \forall n \in \mathbb{Z}$ I'm reading new alien mathematics here, whats is this

You forgot that $\arccos(\frac{\sqrt{3}}{2}) =2\pi n \pm \frac{\pi}{6}, \forall n \in \mathbb{Z}$

It's not quite true as $0\leq \arccos x \leq \pi /2$.