How do you find the eccentricity, directrix, focus and classify the conic section #r=0.8/(1-0.8sintheta)#?

1 Answer
Jul 17, 2017

Answer:

Please see below.

Explanation:

It is a typical equation of an ellipse in polar form. However, it is easier to identify conic section, its eccentricity, directrix and focus in rectangular coordinates. Hence, let us convert the polar equation in rectangular form.

The relation between polar form #(r,theta)# and rectangular form #(x,y)# is given by #x=rcostheta# and #y=rsintheta# i.e. #r^2=x^2+y^2#.

Hence #r=0.8/(1-0.8sintheta)# can be written as

#r-4/5rsintheta=4/5# or #5r-4rsintheta=4#

or #5sqrt(x^2+y^2)-4y=4#

or #25x^2+25y^2=(4+4y)^2=16y^2+32y+16#

or #25x^2+9y^2-32y-16=0#

or #25x^2+9(y^2-2xx16/9y+(16/9)^2)-256/9-16=0#

or #25x^2+9(y-16/9)^2=400/9#

or #x^2/(400/(9xx25))+(y-16/9)^2/(400/(9xx9))=1#

or #x^2/(4/3)^2+(y-16/9)^2/(20/9)^2=1#

Hence, this is the equation of an ellipse of the form #(x-h)^2/a^2+(y-k)^2/b^2=1#,

whose center is #(0,16/9)#, major axis parallel to #y#-axis is #2xx20/9=40/9# and minor axis parallel to #x#-axis is #2xx4/3=8/3#

eccentricity is given by #e=sqrt(1-a^2/b^2)#

= #sqrt(1-(4/3)^2/(20/9)^2)=sqrt(1-9/25)=0.8#

Focii are #(h,k+-be)# i.e. #(0,16/9+-16/9)# i.e. #(0,0)# and #(0,32/9)#

and directrix are #y=k+-b/e# i.e. #y=16/9+-25/9#

i.e. #y=41/9# and #y=-1#

graph{(25x^2+9y^2-32y-16)(x^2+y^2-0.01)(x^2+(y-32/9)^2-0.01)(y+1)(y-41/9)=0 [-6.31, 6.346, -1.44, 4.884]}