Answer
a)-d) See proofs
Work Step by Step
We are given the equation:
$Ax^2+Dx+Ey+F=0$, $A\not=0$
a) Rewrite the equation:
$A\left(x^2+\dfrac{D}{A}x\right)+Ey+F=0$
$A\left(x^2+2\cdot\dfrac{D}{2A}x+\dfrac{D^2}{4A^2}\right)-\dfrac{D^2}{4A}+Ey+F=0$
$A\left(x+\dfrac{D}{2A}\right)^2=-Ey-F+\dfrac{D^2}{4A}$
$\left(x+\dfrac{D}{2A}\right)^2=-\dfrac{E}{A}y-\dfrac{F}{A}+\dfrac{D^2}{4A^2}$
$\left(x+\dfrac{D}{2A}\right)^2=-\dfrac{E}{A}y+\dfrac{D^2-4AF}{4A^2}$
$\left(x+\dfrac{D}{2A}\right)^2=-\dfrac{E}{A}\left(y-\dfrac{D^2-4AF}{4EA^2}\right)$
The equation describes a vertical parabola.
b) We have:
$\left(x+\dfrac{D}{2A}\right)^2=-\dfrac{E}{A}y+\dfrac{D^2-4AF}{4A^2}$
When $E=0$ and $D^2-4AF=0$, we get:
$\left(x+\dfrac{D}{2A}\right)^2=-\dfrac{0}{A}y+\dfrac{0}{4A^2}$
$\left(x+\dfrac{D}{2A}\right)^2=0$
$x+\dfrac{D}{2A}=0$
$x=-\dfrac{D}{2A}$
Therefore in this case the equation represents a vertical line.
c) When $E=0$ and $D^2-4AF>0$ we get:
$\left(x+\dfrac{D}{2A}\right)^2=-\dfrac{0}{A}y+\dfrac{D^2-4AF}{4A^2}$
$\left(x+\dfrac{D}{2A}\right)^2=\dfrac{D^2-4AF}{4A^2}$
$x+\dfrac{D}{2A}=\pm\sqrt{\dfrac{D^2-4AF}{4A^2}}$
Therefore, in this case the equation represents two vertical lines:
$x=-\dfrac{D}{2A}-\sqrt{\dfrac{D^2-4AF}{4A^2}}$
$x=-\dfrac{D}{2A}+\sqrt{\dfrac{D^2-4AF}{4A^2}}$
d) When $E=0$ and $D^2-4AF<0$, we get:
$\left(x+\dfrac{D}{2A}\right)^2=-\dfrac{0}{A}y+\dfrac{D^2-4AF}{4A^2}$
$\left(x+\dfrac{D}{2A}\right)^2=\dfrac{D^2-4AF}{4A^2}$
$\left(x+\dfrac{D}{2A}\right)^2<0$
Therefore, in this case there is no point to check the equation.
