Chapter A5.1, Problem 31E

To determine

To find: conic section equation, vertices, foci, directrices, center, and asymptotes

Answer to Problem 31E

Vertices are $[(-1,2)$ and (-1,0)]

Foci are $[(-1,1\pm \sqrt{2})]$

Center is [(-1,1)]

Asymptotes are $y=\pm (x+1)+1$

Explanation of Solution

Given:

  $y^2-x^2=1$ right 1, up 1

Calculation:

The new parabola is:

The given new hyperbola is:

  ${(y-1)}^2-{(x+1)}^2=1$

Comparing with the standard form:

  $\frac{Y^2}{a^2}-\frac{X^2}{b^2}=1$

Thus,

  $a=1$

  $b=1$

  $X=x+1$

  $Y=y-1$

Therefore

  $c=\sqrt{a^2+b^2}$

  $=\sqrt{1+1}$

  $=\sqrt{2}$

Now, Vertices are given by $(X,Y)=(0,\pm a),$ or $(x+1,y-1)=(0,\pm 1)$

Therefore vertices are $[(-1,2)$ and (-1,0)]

Foci are given by $(X,Y)=(0,\pm c),$ or $(x+1,y-1)=(0,\pm \sqrt{2})$

Therefore foci are $[(-1,1\pm \sqrt{2})]$

Center is given by $(X,Y)=(0,0),$ or $(x+1,y-1)=(0,0)$

Therefore center is [(-1,1)]

Asymptotes are given by $Y=\pm \frac{a}{b}X$ , or $y=\pm (x+1)+1$

Conclusion:

Therefore Vertices are $[(-1,2)$ and (-1,0)]

Foci are $[(-1,1\pm \sqrt{2})]$

Center is [(-1,1)]

Asymptotes are $y=\pm (x+1)+1$