Chapter 5.6, Problem 46HP

(a)

To determine

To show:

The equation of the parabola is of the given form.

Explanation of Solution

Given:

The equation of parabola

  $y=a{x}^2+bxy+c{y}^2+dx+ey+f=0$

Vertical line if $e=0,b^2-4ac=0$

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 $b^2-4ac=0$

  $\begin{array}{l}a{x}^2+bxy+c{y}^2+dx+ey+f=0\\ a{x}^2+dx+0+f=0\\ a{x}^2+dx+f=0\end{array}$

So the quadratic form is

  $b^2-4ac=0$

Therefore a vertical line.

(b)

To determine

To show:

The quadratic function of the parabola is of the given form

Explanation of Solution

Given:

The equation of parabola

  $y=a{x}^2+bxy+c{y}^2+dx+ey+f=0$

Vertical line if $e=0and{b}^2-4ac>0$

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 $e=0and{b}^2-4ac>0$

  $\begin{array}{l}y=a{x}^2+bxy+c{y}^2+dx+ey+f=0\\ a{x}^2+dx+0+f=0\\ a{x}^2+dx+f=0\end{array}$

So the quadratic form is

  $b^2-4ac>0$

Therefore two vertical line

(d)

To determine

To show:

The graph of the parabola is of the given form

Explanation of Solution

Given:

The equation of parabola

  $y=a{x}^2+bxy+c{y}^2+dx+ey+f=0$

Vertical line if $e=0and{b}^2-4ac<0$

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 $e=0and{d}^2-4ac<0$

  $\begin{array}{l}y=a{x}^2+bxy+c{y}^2+dx+ey+f=0\\ a{x}^2+dx+0+f=0\\ x^2+dx+f=0\end{array}$

So the quadratic form is

  $b^2-4ac<0$

Therefore no points of the parabola

(e)

To determine

To show:

The equation of the parabola is of the given form.

Explanation of Solution

Given:

The equation of parabola

  $y=a{x}^2+bxy+c{y}^2+dx+ey+f=0$

Vertical line if $e=0,b^2-4ac=0$

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 $b^2-4ac=0$

  $\begin{array}{l}a{x}^2+bxy+c{y}^2+dx+ey+f=0\\ a{x}^2+dx+0+f=0\\ a{x}^2+dx+f=0\end{array}$

So the quadratic form is

  $b^2-4ac=0$

  $\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Therefore a vertical line.