conjugate axis on the x-axis, one focus at (0, √13), equation of one directrix y= 9√13 / 13. HELP PLEASE

Question

anonymous · Answer

what you need equation of parabola?

anonymous · Answer

hyperbola..

anonymous · Answer

since x- axis is your conjugate you get the two conics one on top and one in the bottom

anonymous · Answer

given eccentricity,e=9/sqrt13
b^2=a^2(e^2-1)

anonymous · Answer

no background about conics.. im sorry

anonymous · Answer

standard form is (x/a)^2-(y/b)^2=1
therefore your equation will be of the form (bx)^2-(ay)^2=(ab)^2

anonymous · Answer

yes yes..

anonymous · Answer

foci: (h, k + c), (h, k - c)
c=sqrt(a^2+b^2)
(h,k) will be your vertex
equation is (y-k)^2/a^2-((x-h)/b)^2=1

anonymous · Answer

there is a change in slope (negative ) because x axis is the conjugate axis

anonymous · Answer

it would be imparted with parabola. i think?