Chapter 11, Problem 41RE

To determine

To find: The type of curve; its foci, vertices and plot the relevant graph.

Answer to Problem 41RE

So given equation is an Ellipse

But negative constant doesn’t form an ellipse

So, there is no focus and no vertices. It is impossible to draw graph.

Explanation of Solution

Given information:

The equation is $9x^2+8y^2-15x+8y+27=0$ .

Calculation:

  $\text{For a Quadratic equation; }a{x}^2+bxy+c{y}^2+dx+ey+f=0.$

  $\begin{array}{l}\text{if }{b}^2-4ac=0\to \text{ it is parabola}\text{.}\\ \text{if }{b}^2-4ac\text{>0}\to \text{ it is hyperbola}\text{.}\end{array}$

  $\begin{array}{l}\text{if }{b}^2-4ac<0\text{ and }b=0\text{ \& }a=c\to \text{ it is circle}\text{.}\\ \text{if }{b}^2-4ac<0\text{ and }b\ne 0\&a\ne c\to \text{it is ellipse}\text{.}\end{array}$

Given equation is; $9x^2+8y^2-15x+8y+27=0$

  $\text{Here; }a=9,b=0,c=8$

  $\begin{array}{l}\text{Now; }{b}^2-4ac={\left(0\right)}^2-4\left(9\right)\left(8\right)\\ =\text{ -28}8<0\end{array}$

So given equation is an Ellipse.

Given equation is $\begin{array}{l}9x^2-15x+8y^2+8y+27=0\\ 9(x^2-\frac{5}{3}x)+8(y^2+y)+27=0\\ 9{(}^x+8{(}^y=\frac{-75}{4}\end{array}$

But negative constant doesn’t form an ellipse

So, no focus and no vertices. It is impossible to draw graph.

Interpretation :

Given curve is theoretically ellipse but doesn’t exist.

  $\begin{array}{l}\text{The foci= no focus}\\ \text{Vertices}=\text{ no vertices}\end{array}$