Chapter 10.5, Problem 56AYU

(A)

To determine

To show:

The equation of the parabola is of the given form

Explanation of Solution

Given:

The equation of parabola

  $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$

Vertical line if $E=0and{D}^2-4AF=0$

Concept used:

The stabdard form of parabola equation

  $y=a{x}^2+bx+c$

Putting $x=h,y=k$ in equation (1)

  $k=a{h}^2+bh+c\mathrm{.......................}\left(2\right)$

Where $\left(h,k\right)$ is the vertex

Calculation:

The stabdard form of parabola equation

  $y=a{x}^2+bx+c$

Putting $x=h,y=k$ in equation (1)

  $\begin{array}{l}k=a{h}^2+bh+c\\ {\left(y-k\right)}^2=4a\left(x-h\right)\mathrm{.......................}\left(2\right)\end{array}$

Where $\left(h,k\right)$ is the vertex

Vertical line if $E=0and{D}^2-4AF=0$

  $\begin{array}{l}A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0\\ A{x}^2+Dx+0+F=0\\ A{x}^2+Dx+F=0\end{array}$

So the quadratic form is $D^2-4AF=0$

Therefore a vertical line

(B)

To determine

To show:

The graph of the parabola is of the given form

Explanation of Solution

Given:

The equation of parabola

  $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$

Vertical line if $E=0and{D}^2-4AF>0$

Concept used:

The stabdard form of parabola equation

  $y=a{x}^2+bx+c$

Putting $x=h,y=k$ in equation (1)

  $k=a{h}^2+bh+c\mathrm{.......................}\left(2\right)$

Where $\left(h,k\right)$ is the vertex

Calculation:

The stabdard form of parabola equation

  $y=a{x}^2+bx+c$

Putting $x=h,y=k$ in equation (1)

  $\begin{array}{l}k=a{h}^2+bh+c\\ {\left(y-k\right)}^2=4a\left(x-h\right)\mathrm{.......................}\left(2\right)\end{array}$

Where $\left(h,k\right)$ is the vertex

Vertical line if $E=0and{D}^2-4AF>0$

  $\begin{array}{l}A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0\\ A{x}^2+Dx+0+F=0\\ A{x}^2+Dx+F=0\end{array}$

So the quadratic form is $D^2-4AF>0$

Therefore two vertical line

(C)

To determine

To show:

The graph of the parabola is of the given form

Explanation of Solution

Given:

The equation of parabola

  $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$

Vertical line if $E=0and{D}^2-4AF<0$

Concept used:

The stabdard form of parabola equation

  $y=a{x}^2+bx+c$

Putting $x=h,y=k$ in equation (1)

  $k=a{h}^2+bh+c\mathrm{.......................}\left(2\right)$

Where $\left(h,k\right)$ is the vertex

Calculation:

The stabdard form of parabola equation

  $y=a{x}^2+bx+c$

Putting $x=h,y=k$ in equation (1)

  $\begin{array}{l}k=a{h}^2+bh+c\\ {\left(y-k\right)}^2=4a\left(x-h\right)\mathrm{.......................}\left(2\right)\end{array}$

Where $\left(h,k\right)$ is the vertex

Vertical line if $E=0and{D}^2-4AF<0$

  $\begin{array}{l}A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0\\ A{x}^2+Dx+0+F=0\\ A{x}^2+Dx+F=0\end{array}$

So the quadratic form is $D^2-4AF<0$

Therefore no points of the parabola