Chapter A5.1, Problem 26E

(a)

To determine

To find: the foci, vertices, and center of the new ellipse.

Answer to Problem 26E

vertices are $[(0,3)$ and (8,3)]

Foci are $(4\pm \sqrt{7},3)$

Center is $(4,3)$

Explanation of Solution

Given:

The ellipse $\frac{{(x)}^2}{16}+\frac{{(y)}^2}{9}=1$

  $\frac{{(x-4)}^2}{16}+\frac{{(y-3)}^2}{9}=1$

Calculation:

The given new ellipse is:

  $\frac{{(x-4)}^2}{16}+\frac{{(y-3)}^2}{9}=1$

Comparing with the standard form:

  $\frac{X^2}{a^2}+\frac{Y^2}{b^2}=1$

Thus,

  $a=4$

  $b=3$

  $X=x-4$

  $Y=y-3$

Therefore

  $c=\sqrt{a^2-b^2}$

  $=\sqrt{7}$

Now, Vertices are given by $(X,Y)=(\pm a,0),$ or $(x-4,y-3)=(\pm 4,0)$

Therefore vertices are $[(0,3)$ and (8,3)]

Focus is given by $(X,Y)=(\pm c,0),$ or $(x-4,y-3)=(\pm \sqrt{7},0)$

Therefore vertices are $[(0,3)$ and (8,3)]

Focus is given by $(X,Y)=(\pm c,0),$ or $(x-4,y-3)=(\pm \sqrt{7},0)$

Therefore foci are $(4\pm \sqrt{7},3)$

Center is given by $(X,Y)=(0,0),$ or $(x-4,y-3)=(0,0)$

Therefore center is $(4,3)$

Conclusion:

Therefore vertices are $[(0,3)$ and (8,3)]

Foci are $(4\pm \sqrt{7},3)$

Center is $(4,3)$