Thread: Conic Sections?

Show algebraically that the graph of an equation of the form:
Ax2 + Dx + Ey + F = 0, A 0

a) Is a parabola if E 0.
b) Is a vertical line if E = 0 and D2 - 4AF = 0.
c) Is two vertical lines if E = 0 and D2 - 4AF > 0.
d) Contains no points if E = 0 and D2 - 4AF < 0.

Originally Posted by tle8807

Show algebraically that the graph of an equation of the form:
Ax2 + Dx + Ey + F = 0, A 0

a) Is a parabola if E 0.

if $\displaystyle E \neq 0$, then we can write the equation in the form of $\displaystyle Y = - \, \frac{Ax^2 + Dx + F}{E}$ you're done. but if you want the form $\displaystyle y=4p(x-h)$, just factor the numerator by completing the square.

Originally Posted by tle8807

b) Is a vertical line if E = 0 and D2 - 4AF = 0.

If E = 0, then you have the quadratic equation $\displaystyle Ax^2 + Dx + F = 0$
note that if you use the quadratic formula in solving x, $\displaystyle D^2 - 4AF$ is the determinant, and if it is zero, you will be left with constant $\displaystyle \frac{-D}{2A}$

Originally Posted by tle8807

c) Is two vertical lines if E = 0 and D2 - 4AF > 0.

same with b). but here, note that for a quadratic equation in one variable, if the discriminant is positive, then you have two solutions for the variable, hence, you have two vertical lines.

Originally Posted by tle8807

d) Contains no points if E = 0 and D2 - 4AF < 0.

same reasong with c) but this time, if the discrimiant is negative, the graph does not intersect the x axis (in 2-d), hence "no" points. (i have to include the quotation marks..)