How do you write the hyperbola #16y^2-36x^2+9=0# in standard form? Precalculus Geometry of a Hyperbola Standard Form of the Equation 1 Answer Ratnaker Mehta Aug 30, 2016 #x^2/4-y^2/(9/16)=1#. Explanation: We rewrite the eqn. as #36x^2-16y^2=9#. #:. (36/9)x^2-(16/9)y^2=1#. #:. x^2/4-y^2/(9/16)=1#. Answer link Related questions What is the standard form of the equation of a hyperbola? What conic section is represented by the equation #(y-2)^2/16-x^2/4=1#? What conic section is represented by the equation #y^2/9-x^2/16=1#? What conic section is represented by the equation #x^2/9-y^2/4=1#? How can I tell the equation of a hyperbola from the equation of an ellipse? What does the equation #9y^2-4x^2=36# tell me about its hyperbola? What does the equation #(x+2)^2/4-(y+1)^2/16=1# tell me about its hyperbola? What does the equation #(x-1)^2/4-(y+2)^2/9=1# tell me about its hyperbola? Why is a hyperbola considered a conic section? How do you write the equation of a hyperbola in standard form given Foci: (3,+-2) and... See all questions in Standard Form of the Equation Impact of this question 2349 views around the world You can reuse this answer Creative Commons License