Chapter A5.1, Problem 34E

To determine

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

Answer to Problem 34E

Vertex is [(-1,1)]

Focus is [(-1,0)]

Explanation of Solution

Given:

  $x^2+2x-4y-3=0$

Calculation:

The given curve is:

  $x^2+2x-4y-3=0$

Converting it into standard form:

  ${(x+1)}^2=-4(y-1)$

Comparing with the standard form for a Parabola:

  $X^2=-4pY$

We get,

  $p=1$

  $X=x+1$

  $Y=y-1$

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

Therefore vertex is [(-1,1)]

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

Therefore focus is [(-1,0)]

Conclusion:

Therefore vertex is [(-1,1)] and focus is [(-1,0)]