Chapter EP, Problem 10.6.16EP

To determine

To check: Whether the graph of each equation is a parabola, circle, ellipse or hyperbola.

Answer to Problem 10.6.16EP

The equation $16{\left(x-3\right)}^2+81{\left(y+4\right)}^2=1296$ represents a Ellipse.

Explanation of Solution

Given:

The equation is $16{\left(x-3\right)}^2+81{\left(y+4\right)}^2=1296$ .

Calculation:

Rewrite the equation as follows.

$\begin{array}{c}16{\left(x^2-6x+9\right)}^2+81{\left(y^2+8y+16\right)}^2=1296\\ 16x^2-96x+144+81y^2+648y+1296=1296\\ 16x^2+81y^2-96x+648y+144=0\end{array}$

Compare the equation $16x^2+81y^2-96x+648y+144=0$ with standard equation $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$ .

$\begin{array}{c}A=16\\ B=0\\ C=81\\ D=-96\end{array}$

And,

$\begin{array}{c}E=648\\ F=144\end{array}$

Substitute all these values in the equation $D=B^2-4AC$ .

$\begin{array}{c}D={\left(0\right)}^2-4\left(16\right)\left(81\right)\\ =-5184\\ <0\end{array}$

Since, the $D<0$ so, the equation $16{\left(x-3\right)}^2+81{\left(y+4\right)}^2=1296$ represents a Ellipse.