Normal to a Hyperbola at Five Points Passes Through Origin I know that a general equation of the second degree represents a conic or a pair of straight lines. I evaluated the discriminant $\Delta$ which turns out to be equal to $ac-b^2$ which from the hypothesis is positive.

From this I concluded that the conic is an ellipse. However I've read that from any point, a maximum of four real normals can be drawn to an ellipse, yet the hypothesis states that five normals to the ellipse can be drawn from the origin. I've got no clue how to proceed.

Edit- To add to my confusion, the ellipse also is centred at origin with its axis inclined to the x,y axes. I simply don't see how there can be an odd number of normals from the centre to an ellipse to it as by symmetry I'd expect an even number of normals.

• It says $ac > b^2$, so $ac - b^2 > 0$... – Dan Uznanski May 14 '17 at 2:33
• @Dan sorry about that, it's an ellipse, but my problem still holds :( – Tim The Enchanter May 14 '17 at 2:47
• Perhaps the ellipse is a circle? – Blue May 14 '17 at 2:54
• @Blue that would seem to be the only option, but the problem is a bit ambiguous as to there being five points and exactly five points. I'm going to wait to see if anyone else has an idea, and if not I'm going to have to assume it to be at least 5 points. – Tim The Enchanter May 14 '17 at 2:59