Chapter 10.5, Problem 54AYU

To determine

To show:

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 $A\ne 0,C\ne 0,F\ne 0$

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

  $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=\mathrm{0.............}\left(1\right)$

  $A^{\prime }{x^{\prime }}^2+B^{\prime }{x}^{\prime }{y}^{\prime }+C^{\prime }{y^{\prime }}^2+D^{\prime }{x}^{\prime }+E^{\prime }{y}^{\prime }+F^{\prime }=\mathrm{0...........}\left(2\right)$

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

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

From equation (1) and equation (2)

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

So the quadratic form is

  $A+C=A^{\prime }+C^{\prime }$