For Exercise, a. Identify the equation as representing a circle, an ellipse, a hyperbola, or a parabola. b. Graph the curve. c. Identify key features of the graph. That is. If the equation represents a circle, identify the center and radius. If the equation represents an ellipse, identify the center, vertices, endpoints of the minor axis, foci, and eccentricity. If the equation represents a hyperbola, identify the center, vertices, foci, equations of the asymptotes, and eccentricity. If the equation represents a parabola, identify the vertex, focus, endpoints of the latus rectum, equation of the directrix, and equation of the axis of symmetry. x2 + y2 − 4x − 6y + 1=0

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For Exercise,  a. Identify the equation as representing a circle, an ellipse, a hyperbola, or a parabola.  b. Graph the curve. c. Identify key features of the graph. That is. If the equation represents a circle, identify the center and radius.  If the equation represents an ellipse, identify the center, vertices, endpoints of the minor axis, foci, and eccentricity.  If the equation represents a hyperbola, identify the center, vertices, foci, equations of the asymptotes, and eccentricity.  If the equation represents a parabola, identify the vertex, focus, endpoints of the latus rectum, equation of the directrix, and equation of the axis of symmetry. x2 + y2 − 4x − 6y + 1=0

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Step 1  a)For the given equation simplify it and compare with standard forms of conic sections to determine which conic section or a figure it represents as. x2 − y2 − 4x − 6y + 1=0 x2 − 4x + 4 − 4 − y2 − 6y + 9 − 9 + 1=0 (x2 − 4x + 4) − (y2 + 6y + 9) + 1 + 9 − 4=0 (x − 2)2 − (y + 3)62 + 6=0 (y + 3)2 − (x − 2)2=6 (y + 3)26 − (x − 2)26=1  Which is a hyperbola.  Step 2  Now to find the features of the given hyperbola there are standard forms where these features can be determined. Let us determine centre ‘C’ of this hyperbola by comparing with standard form of such hyperbola as. (y − k)2a2 − (x − h)2b2=1 (y + 3)26 − (x − 2)26=1 C=(h, k) here C=(2, −3)  Step 3  Vertices &quot;V&quot; for this hyperbola can be determined to be as. V=(h, k ± a) a2=6 a= 6 V=(2−3 ± 6)  Step 4  Foci &quot;F&quot; for this hyperbola can be determined to be as F=(h, k ± c) c2=a2 + b2 c2=6 + 6 c= 12 F=(2, −3 ± 12)  Step 5  Equation of asymptotes for this hyperbola can be determined to be as. y= ± ab(x − h) + k y= ± 66(x − 2) − 3 y= ± (x − 2) − 3  Step 6  Eccentricity &quot;e&quot; is calculated by the formula as. e= a2=b2a = 6 + 66 = 126 = 2 66 = 2

For Exercise, a. Identify the equation as representing a circle, an ellipse, a hyperbola, or a parabola. b. Graph the curve. c. Identify key features

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