# Thread: one-parameter family of conics

1. ## one-parameter family of conics

Hi to everybody!!!!

I have to find a solution to these problems, but i'm not able.
The question are:

1.What is the general formula for the one parameter family of conics through the points (1,1), (1,-1), (-1,-1), (-1,1).

I hope someone can help me! Thank you very much.

2. Suppose R=[u] and S=[s] are two points in the projective space neither lying an a given conic defined by the symmetric bilinear form B. Show that RS is tangent to the conic if and only if B(u,u)B(v,v)-(B(u,v))^2=0.

Thanks!

2. Originally Posted by womaninmaths
Hi to everybody!!!!

I have to find a solution to these problems, but i'm not able.
The question are:

1.What is the general formula for the one parameter family of conics through the points (1,1), (1,-1), (-1,-1), (-1,1).

...
Guten Morgen,

there are 3 (one-parameter) families which passes through the vertices of this square:

1. ellipses
2. hyperbolae opening to the right and the left
3. hyperbolae opening up and down

The center of all conics is the origin.

to #1.: The general equation of an ellipse with center at (0, 0) and the axes a and b is:

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ ..........Plug in the coordinates of (-1, 1) and (1, 1):

$\frac{1}{a^2}+\frac{1}{b^2}=1 ~\implies~b^2 = \frac{a^2}{a^2-1}$ .......... Plug in this term for bē into the original equation of the ellipse:

$\frac{x^2}{a^2}+\frac{y^2}{\frac{a^2}{a^2-1}}=1 ~\implies~ \frac{x^2}{a^2}+\frac{(a^2-1) \cdot y^2}{a^2}=1$

Don't forget to calculate the domain of the family!

To demonstrate the effect of this equation I've attached a drawing with 1 < a < 5

To #2, #3: Use the same method.