Conic Sections?

• Nov 26th 2007, 05:04 AM
tle8807
Conic Sections?
Show algebraically that the graph of an equation of the form:
Ax2 + Dx + Ey + F = 0, A http://www.alpertron.com.ar/NOTEQUAL.GIF0

a) Is a parabola if E http://www.alpertron.com.ar/NOTEQUAL.GIF0.
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.
• Nov 26th 2007, 05:15 AM
kalagota
Quote:

Originally Posted by tle8807
Show algebraically that the graph of an equation of the form:
Ax2 + Dx + Ey + F = 0, A http://www.alpertron.com.ar/NOTEQUAL.GIF0

a) Is a parabola if E http://www.alpertron.com.ar/NOTEQUAL.GIF0.

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

Quote:

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

Quote:

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.

Quote:

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..)