Chapter 10.2, Problem 79AYU

Show that the graph of an equation of the form $C{y}^2\text{ + }Dx\text{ + }Ey\text{ + }F\text{ = 0}$     $C\ne \text{ 0}$

(a) Is a parabola if $D\ne \text{ 0}$.

(b) Is a horizontal line if $D=\text{ 0}$ and $E^2-\text{ 4}CF\text{ = 0}$.

(c) Is two horizontal lines if $D=\text{ 0}$ and $E^2-\text{ 4}CF\text{ > 0}$.

(d) Contains no points if $D=\text{ 0}$ and $E^2-\text{ 4}CF\text{ < 0}$.

To determine

To find: An equation of the form $Cy²+Dx+Ey+F=0$; $C\ne 0$.

a. Is a parabola if $D\ne 0$.

Answer to Problem 79AYU

A. This is the equation of the parabola whose vertex is $(\frac{E^2-4FC}{4DC},-\frac{E}{2C})$ and the axis of symmetry is parallel to $x\text{-axis}$.

Explanation of Solution

Given:

An equation of the form $Cy²+Dx+Ey+F=0$; $C\ne 0$.

Formula used:

Vertex	Focus	Directrix	Equation
$\left(h,k\right)$	$\left(h+a,k\right)$	$x=h-a$	$\left(y-k\right)²=4a\left(x-h\right)$

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Calculation:

a. If $\text{D }\ne 0$,

$Cy²+Dx+Ey+F=0$

$C{y}^2+Ey=-Dx-F$

Adding $\frac{E^2}{4C}$ both sides,

$C\left(y^2+\frac{E}{C}y+\frac{E^2}{4C^2}\right)=-Dx-F+\frac{E^2}{4C}$

$C{\left(y+\frac{E}{2C}\right)}^2=-Dx-F+\frac{E^2}{4C}$

${\left(y+\frac{E}{2C}\right)}^2=-\frac{D}{C}\left(x-\frac{F}{D}+\frac{E^2}{4DC}\right)$

${\left(y+\frac{E}{2C}\right)}^2=-\frac{D}{C}\left(x-\frac{E^2-4FC}{4DC}\right)$

This is the equation of the parabola whose vertex is $(\frac{E^2-4FC}{4DC},-\frac{E}{2C})$ and the axis of symmetry is parallel to $x\text{-axis}$.

To determine

To find: An equation of the form $Cy²+Dx+Ey+F=0$; $C\ne 0$.

a. Is a horizontal line if $D=0$ and $E²-4CF=0$.

Answer to Problem 79AYU

It is a quadratic equation; $y=\frac{-E\pm \sqrt{E^2-4CF}}{2C}$ if $E²=4CF$, then $y=-E/2C$ is a single horizontal line.

Explanation of Solution

Given:

An equation of the form $Cy²+Dx+Ey+F=0$; $C\ne 0$.

Formula used:

Vertex	Focus	Directrix	Equation
$\left(h,k\right)$	$\left(h+a,k\right)$	$x=h-a$	$\left(y-k\right)²=4a\left(x-h\right)$

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Calculation:

b. If $D=0$

$Cy²+Dx+Ey+F=0$

$Cy²+Ey+F=0$

It is a quadratic equation; $y=\frac{-E\pm \sqrt{E^2-4CF}}{2C}$ if $E²=4CF$, then $y=-E/2C$ is a single horizontal line.

To determine

To find: An equation of the form $Cy²+Dx+Ey+F=0$; $C\ne 0$.

c. Is a horizontal line if $D=0$ and $E²-4CF>0$.

Answer to Problem 79AYU

c. It is a quadratic equation; $y=\frac{-E\pm \sqrt{E^2-4CF}}{2C}$ if $E²>4CF$, then $y=\frac{-E+\sqrt{E^2-4CF}}{2C}$, $y=\frac{-E-\sqrt{E^2-4CF}}{2C}$ are two horizontal lines.

Explanation of Solution

Given:

An equation of the form $Cy²+Dx+Ey+F=0$; $C\ne 0$.

Formula used:

Vertex	Focus	Directrix	Equation
$\left(h,k\right)$	$\left(h+a,k\right)$	$x=h-a$	$\left(y-k\right)²=4a\left(x-h\right)$

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Calculation:

c. If $D=0$, then

$Cy²+Dx+Ey+F=0$

$Cy²+Ey+F=0$

It is a quadratic equation; $y=\frac{-E\pm \sqrt{E^2-4CF}}{2C}$ if $E²>4CF$, then $y=\frac{-E+\sqrt{E^2-4CF}}{2C}$, $y=\frac{-E-\sqrt{E^2-4CF}}{2C}$ are two horizontal lines.

To determine

To find: An equation of the form $Cy²+Dx+Ey+F=0$; $C\ne 0$.

d. Contains no points if $D=0$ and $E²-4CF<0$.

Answer to Problem 79AYU

It is a quadratic equation; $y=\frac{-E\pm \sqrt{E^2-4CF}}{2C}$ if $E²<4CF$, then there is no real solution for $y$ and hence, the graph contains no points.

Explanation of Solution

Given:

An equation of the form $Cy²+Dx+Ey+F=0$; $C\ne 0$.

Formula used:

Vertex	Focus	Directrix	Equation
$\left(h,k\right)$	$\left(h+a,k\right)$	$x=h-a$	$\left(y-k\right)²=4a\left(x-h\right)$

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Calculation:

d. If $D=0$, then

$Cy²+Dx+Ey+F=0$

$Cy²+Ey+F=0$

It is a quadratic equation; $y=\frac{-E\pm \sqrt{E^2-4CF}}{2C}$ if $E²<4CF$, then there is no real solution for $y$ and hence, the graph contains no points.