Chapter A5.1, Problem 27E

(a)

To determine

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

Answer to Problem 27E

Therefore vertices are $[(-2,0)$ and (6,0)]

Foci are $[(-3,0)$ and (7,0)]

Center is [(2,0)]

Explanation of Solution

Given:

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

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

Calculation:

The given new hyperbola is:

  

Comparing with the standard form:

  $\frac{X^2}{a^2}-\frac{Y^2}{b^2}=1$

Thus,

  $a=4$

  $b=3$

  $X=x-2$

  $Y=y$

Therefore

  $c=\sqrt{a^2+b^2}$

  $=\sqrt{25}$

  $=5$

Now, Vertices are given by $(X,Y)=(\pm a,0)$ , or $(x-2,y)=(\pm 4,0)$

Therefore vertices are

and (6,0)]

Foci are given by $(X,Y)=(\pm c,0),$ or $(x-2,y)=(\pm 5,0)$

Therefore foci are

and (7,0)]

Center is given by $(X,Y)=(0,0),$ or $(x-2,y)=(0,0)$

Therefore center is [(2,0)]

Asymptotes are given by $Y=\pm \frac{b}{a}X$ , or $\left[y=\pm \frac{3}{4}(x-2)\right]$

Conclusion:

Therefore vertices are

and (6,0)]

Foci are

and (7,0)]

Center is [(2,0)]