Chapter A5.1, Problem 35E

To determine

To find: center, foci, vertices, asymptotes and radius.

Answer to Problem 35E

Vertices are $(-2\pm \sqrt{5},0)$ .

Foci are $(-4,0)$ and (0,0).

Center is $(-2,0)$ .

Explanation of Solution

Given:

  $x^2+5y^2+4x=1$

Calculation:

The given curve is:

  $x^2+5y^2+4x=1$

Converting it into standard form:

  $\frac{{(x+2)}^2}{5}+y^2=1$

Comparing with the standard form for an Ellipse:

  $\frac{X^2}{a^2}+\frac{Y^2}{b^2}=1$

Thus,

  $a=\sqrt{5}$

  $b=1$

  $X=x+2$

  $Y=y$

Therefore

  $\begin{array}{l}c=\sqrt{a^2-b^2}\\ =\sqrt{5-1}\\ =2\end{array}$

Now, Vertices are given by $(X,Y)=(\pm a,0),$ or $(x+2,y)=(\pm \sqrt{5},0)$ .

Therefore vertices are $(-2\pm \sqrt{5},0)$ .

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

Therefore foci are $(-4,0)$ and (0, 0).

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

Therefore center is $(-2,0)$ .

Conclusion:

Therefore Vertices are $(-2\pm \sqrt{5},0)$ .

Foci are $(-4,0)$ and (0,0).

Center is $(-2,0)$ .