How do you convert r=2sec^2(theta/2) into cartesian form?

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Answer 1 · 2018-04-20T23:19:20

The Cartesian equation is #y^2=-8(x-2)#.

To convert the equation, use the reciprocal definition:

#sectheta=1/costheta#

and the cosine half-angle formula:

#cos(theta/2)=+-sqrt((1+costheta)/2)#

Here's the equation:

#r=2sec^2(theta/2)#

#r=2*1/cos^2(theta/2)#

#r=2*1/(cos(theta/2))^2#

#r=2*1/(+-sqrt((1+costheta)/2))^2#

#r=2*1/((1+costheta)/2)#

#r=4/(1+costheta)#

#r+rcostheta=4#

Using the substitutions #r=sqrt(x^2+y^2)# and #rcostheta=x#:

#sqrt(x^2+y^2)+x=4#

#sqrt(x^2+y^2)=4-x#

#x^2+y^2=(4-x)^2#

#color(red)cancelcolor(black)(x^2)+y^2=16-8x+color(red)cancelcolor(black)(x^2)#

#y^2=-8x+16#

#y^2=-8(x-2)#

That's the equation; it's leftward-opening parabola. Here's the graph:

graph{y^2=-8(x-2) [-25.66, 25.65, -12.83, 12.83]}

Hope this helped!