Question 1172211
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Case 1: If the transverse axis is parallel to the x-axis (branches open right and left), then the equation is<br>
{{{(x-h)^2/a^2-(y-k)^2/b^2 = 1}}}<br>
Case 2: If the transverse axis is parallel to the y-axis (branches open up and down), then the equation is<br>
{{{(y-k)^2/a^2-(x-h)^2/b^2 = 1}}}<br>
In either equation, (h,k) is the center of the hyperbola; 2a is the length of the transverse axis, and 2b is the length of the conjugate axis.<br>
c is the distance from the center to either focus; a, b, and c are related by<br>
{{{c^2 = a^2+b^2}}}<br>
You need to complete the square in both x and y to put the given equation into one of those forms.<br>
{{{3x^2-2y^2+6x-8y=6}}}
{{{(3x^2+6x)-(2y^2+8y)=6}}}
{{{3(x^2+2x)-2(y^2+4y)=6}}}
{{{3(x^2+2x+1)-2(y^2+4y+4)=6+3-8}}}
{{{3(x+1)^2-2(y+2)^2=1}}}
{{{(x+1)^2/(1/3)-(y+2)^2/(1/2)=1}}}<br>
That is in the form of case 1: (h,k) = (-1,-2); a^2=1/3; b^2=1/2.<br>
Transverse Axis parallel to: x-axis or y-axis? See the definition of case 1.<br>
Graph: (I leave that to you)<br>
Center of the Hyperbola: (h,k)<br>
Vertices of the Hyperbola: a units to the right and left of the center<br>
Foci of the Hyperbola: c units to the right and left of the center<br>
Equations of the Asymptotes: the slopes are b/a and -b/a; (h,k) is a point on both asymptotes.<br>