How do you identify the vertices, foci, and direction of #x^2/20-(y+1)^2/10=1#?

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Answer 1 · 2017-10-15T18:10:55

When given an equation of the form:

#(x-h)^2/a^2-(y-k)^2/b^2=1 " [1]"#

The negative y term indicates a horizontal transverse axis.

The vertices are #(h-a,k)# and #(h+a,k)#

The foci are #(h-sqrt(a^2+b^2),k)# and #(h+sqrt(a^2+b^2),k)#

With very little work, we can write the given equation:

#x^2/20-(y+1)^2/10=1#

In the form of equation [1]:

#(x- 0)^2/(2sqrt5)^2-(y-(-1))^2/(sqrt10)^2=1#

The negative y term indicates a horizontal transverse axis.

The vertices are #(-2sqrt5,-1)# and #(2sqrt5,-1)#

Do the calculation for #sqrt(a^2+b^2) = sqrt(20+10) = sqrt(30)#

The foci are #(-sqrt(30),-1)# and #(sqrt30,-1)#