Chapter 9.3, Problem 100E

To determine

To prove that the graph of the equation

  $\text{A}{x}^2+C{y}^2+Dx+Ey+F=0$ is one of the following (except in degenerate cases)

  

Answer to Problem 100E

(b) Parabola is the graph of the equation $\text{A}{x}^2+C{y}^2+Dx+Ey+F=0$

Explanation of Solution

Given information : 

The equation $\text{A}{x}^2+C{y}^2+Dx+Ey+F=0$ and

  

Concept used:

  $\begin{array}{l}\text{The graph of A}{x}^2+C{y}^2+Dx+Ey+F=0\text{ is one of the following}\\ 1.\text{ Circle A=C }\left(\text{A}\ne \text{0}\right)\\ 2.\text{ Parabola AC=0 }\left(\text{A=0 or C=0,but not both}\right)\\ 3.\text{ Ellipse AC>0 }\left(\text{A and C have like signs}\right)\\ \text{4}\text{. Hyperbola AC<0 }\left(\text{A and C have unlike signs}\right)\end{array}$

Explanation: 

Clearly conditions are completely given only in case of (b) parabola

So, proving $\text{A}{x}^2+C{y}^2+Dx+Ey+F=0$ is graph of Parabola

Let A=1 and C=0

Putting these value in equation

  $\begin{array}{l}\text{A}{x}^2+C{y}^2+Dx+Ey+F=0\\ x^2+Dx+Ey+F=0\\ x^2+Dx=-Ey-F\\ \text{completing square (on left hand side)}\\ x^2+Dx+\frac{D^2}{4}=-Ey-F+\frac{D^2}{4}\\ {\left(x+\frac{D}{2}\right)}^2=-Ey-F+\frac{D^2}{4}\\ {\left(x-\left(\frac{-D}{2}\right)\right)}^2=-E\left(y-\left(-\frac{F}{E}-\frac{D^2}{4E}\right)\right)\\ put\left(\frac{-D}{2}\right)=h,-E=4a,\left(-\frac{F}{E}-\frac{D^2}{4E}\right)=k\\ {\left(x-h\right)}^2=4a\left(y-k\right)\\ The\text{ derived equation is a parabola in Standard form}\end{array}$