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