Graph the equation. #(3x^2)-(2y^2)-9x+4y-8=0# What are all applicable points (vertex, focus, center, etc)?

1 Answer
A. S. Adikesavan
Nov 21, 2016

Hyperbola: center C(1.5, 1), semi major axis a = 2.062, semi transverse axis b = 2.525, eccentricity , vertices A(3.562, 1), and A'(#-0.562#, 1), e = 1.581, and foci S(4.760, 1) and S'(-0.760, 1),

Explanation:

The equation can be reorganized to the standard form

#(x-3/2)^2/(sqrt(17/4))^2-(y-1)^2/(sqrt(51/8))^2=1# revealing

Center C(1.5, 1)

Major axis y = 1

Transverse axis x = 1.5

Semi major axis a = sqrt(17/4)=2.062

Vertices on major axis, distant a from C:

Vertices A(3.562, 1), and A'(#-0.562#,

Semi transverse axis b = sqrt(51/8) = 2.525

Eccentricity e = sqrt(1+b^2/a^2) = 1.581,

Foci on the major axis, distant ae from C#

Foci S(4.760, 1) and S'(-1.760, 1)

Directrices parallel to transverse axis, distant a/e from it:

Directrrices x = 2.804 and x = 0.196

graph{3x^2-2y^2-9x+4y-8=0 [-10, 10, -5, 5]}

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