anonymous
  • anonymous
conjugate axis on the x-axis, one focus at (0, √13), equation of one directrix y= 9√13 / 13. HELP PLEASE
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
what you need equation of parabola?
anonymous
  • anonymous
hyperbola..
anonymous
  • anonymous
since x- axis is your conjugate you get the two conics one on top and one in the bottom

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anonymous
  • anonymous
given eccentricity,e=9/sqrt13 b^2=a^2(e^2-1)
anonymous
  • anonymous
no background about conics.. im sorry
anonymous
  • anonymous
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
  • anonymous
yes yes..
anonymous
  • anonymous
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
  • anonymous
there is a change in slope (negative ) because x axis is the conjugate axis
anonymous
  • anonymous
it would be imparted with parabola. i think?

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