How do you find the vertices and foci of #4y^2-81x^2=324#? Precalculus Geometry of a Hyperbola General Form of the Equation 1 Answer Nallasivam V Jan 11, 2017 Vertices #(0, 9) ; (0, -9)# Foci #(0, 9.22) ; (0, -9.22)# Explanation: Answer is given as image Answer link Related questions What type of conic section has the equation #4x^2 - 25y^2 - 50y - 125= 0#? What type of conic section has the equation #4x^2 - y^2 +4y - 20= 0#? What type of conic section has the equation #9y^2 - x^2 - 4x+54y +68= 0#? What type of conic section has the equation #4x^2 - y^2 - 16x - 2y+11=0= 0#? How do I convert the equation #4x^2 - y^2 - 24x+4y +28= 0# to standard form? How do I use completing the square to convert the general equation of a hyperbola to standard form? How do I convert the equation #-9x^2 +4y^2 +72x-16y =164# to standard form? What type of conic section has the equation #4x^2-y^2=16#? What type of conic section has the equation #9x^2-4y^2=36#? How do I convert the equation #9x^2−y^2+54x+10y+55=0# to standard form? See all questions in General Form of the Equation Impact of this question 2269 views around the world You can reuse this answer Creative Commons License