1005 MrG What is a Twisted Conic

2592 days ago by MATH4R2013

var('theta') 
       
#1) Graph 5*x^2+6*x*y+5*y^2-8==0 in polar mode! #1) Convert x=r*cos(theta) #1) Convert y=r*sin(theta) #1) Solve for r=f(theta) polar_plot([sqrt(8/(5+6*sin(theta)*cos(theta)))],(theta,0,2*pi),color='red',thickness=4,aspect_ratio=1) 
       
#2) Graph 2*x^2+2*y^2-8*x+8*y==0 #2) Graph 2*(r*cos(theta))^2+2*(r*sin(theta))^2-8*(r*cos(theta))+8*(r*sin(theta))==0 #2) Graph [2*(cos(theta))^2+2*(sin(theta))^2]*r^2+[8*sin(theta)-8*cos(theta)]*r==0 #2) Graph [2*(cos(theta))^2+2*(sin(theta))^2]*r+[8*sin(theta)-8*cos(theta)]==0 #2) Graph 2*r=[8*cos(theta)-8*sin(theta) #2) Graph r=4*cos(theta)-4*sin(theta) polar_plot([4*cos(theta)-4*sin(theta)],(theta,0,2*pi),color='green',thickness=8,aspect_ratio=1) 
       
#2) Graph 2*x^2+2*y^2-8*x+8*y==0 #2) Graph 2*x^2-8*x+2*y^2+8*y==0 #2) Graph 2*(x^2-4*x)+2*(y^2+4*y)==0 #2) Graph 2*(x^2-4*x+4)+2*(y^2+4*y+4)==0+8+8 #2) Graph 2*(x-2)^2+2*(y+2)^2==16 #2) Graph (x-2)^2/8+(y+2)^2/8==1 #2) Graph (y+2)^2/8==1-(x-2)^2/8 #2) Graph (y+2)^2==8-(x-2)^2 #2) Graph (y+2)==+/-sqrt(8-(x-2)^2) #2) Graph y==-2+/-sqrt(8-(x-2)^2) plot([-2-sqrt(8-(x-2)^2),-2+sqrt(8-(x-2)^2)],2-sqrt(8),2+sqrt(8),color='green',thickness=8,aspect_ratio=1)