How do you find the standard form given #x^2-2x-4y-11=0#? Precalculus Geometry of a Hyperbola Standard Form of the Equation 1 Answer mimi Dec 12, 2017 #y=x^2/4-x/2-11/4# Explanation: STANDARD FORM: #y=ax^2+bx+c# Make #x^2-2x-4y-11=0# look like #y=ax^2+bx+c#. #x^2-2x-4y-11=0# [Add #4# to both sides.] #x^2-2x-11=4y# [Divide by #4# to isolate y.] #x^2/4-(2x)/4-11/4=y# [Simplify.] #x^2/4-x/2-11/4=y# ANSWER: #y=x^2/4-x/2-11/4# 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 2088 views around the world You can reuse this answer Creative Commons License