Instructions:Graph the conic section and make sure to label the coordinates in the graph. Give the standard form (SF) and the general form (GF) of the conic sections.CIRCLES:Center is at (2, −4). the diameter's length is 6. The endpoints of the diameter is at (−1, −4) and (3, 6).

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Instructions:Graph the conic section and make sure to label the coordinates in the graph. Give the standard form (SF) and the general form (GF) of the conic sections.CIRCLES:Center is at (2, −4). the diameter&#039;s length is 6. The endpoints of the diameter is at (−1, −4) and (3, 6).

Aubree Mcintyre · Accepted Answer

Step 1 We have to find standard form and general form of conic section and also we have to make the graph of these conic. Step 2 Centre (2, −4) and diameter&#039;s length is 6. We have given diameter =6 So, radius of circle (r)=62=3 Concept: If centre of a circle is (h, k) and radius of circle is r, then equation of circle in standard form is: (x−h)2+(y−k)2=r2 And we can find general form of circle after expanding standard form: We have centre of circle (2, −4) radius 3 So, equation of circle in standard form is: (x−2)2+(y−(−4))2=32 or (x−2)2+(y+4)2=32 Now, Expland (x−2)2+(y+4)2=32 we get x2+y2−4x+8y+4+16=9 x2+y2−4x+8y+11=0 So, general form of circle is: x2+y2−4x+8y+11=0 Now, graph of this circle:   Step 3 End points of the diameter are (−1, −4) and (3, 6) Centre of circle lies on the mid-point of the diameter So, centre of coordinate (x, y)=(3−12, 6−42)=(1, 1) And distance between (−1, −4) and (3, 6) =(6−(−4))2+(3−(−1))2 =(6+4)2+(3+1)2 =100+16 =116 =229 Distance of diameter =229 So, radius of circle (r)=2292=29 We have centre (1, 1) and radius (r)=29 So, equation of circle in standard form is: (x−1)2+(y−1)2=(29)2 Now, Expand (x−1)2+(y−1)2

Instructions: Graph the conic section and make sure to label the

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