Question

We have ellipse     \( x^2/a^2 +y^2/b^2 =1 \).

Show that the polar equation the ellipse.

 \( x^2/a^2 +y^2/b^2 =1 \) is \( r^2=b^2/(1-e^2cos^2θ) \)

 

Answer 1

Solution:

We know that in polar coordinate \( x=rcos(θ) \) and \( y=rsin(θ) \)

The ellipse equation become

\( =>x^2/a^2+y^2/b^2=1 =>(rcosθ)^2/a^2+(rsinθ)^2/b^2=1 \)

\( =>r^2(cos^2θ/a^2+sin^2θ/b^2)=1 \)

\( =>r^2(b^2cos^2θ+a^2sin^2θ)=a^2b^2 \)

\( =>r^2=a^2b^2/(b^2cos^2θ+a^2sin^2θ) \)

In ellipse we know \( e^2=(a^2-b^2)/a^2 \) . Then add few term to equation.

\( =>r^2=a^2b^2/(b^2cos^2θ+a^2sin^2θ-a^2cos^2θ+a^2cos^2θ) \)

\( =>r^2=a^2b^2/((b^2-a^2)cos^2θ+a^2) \) divide numerator and denominator by \( a^2 \) 

\( =>r^2=b^2/[((b^2-a^2)/a^2)cos^2θ+1] \)

\( =>r^2=b^2/(1-e^2cos^2θ) \)

Therefore, the polar equation of ellipse \( x^2/a^2+y^2/b^2=1 \) is \( r^2=b^2/(1-e^2cos^2θ) \)

 

Therefore, the polar equation of ellipse 

\( x^2/a^2 +y^2/b^2 =1 \) is \( r^2=b^2/(1-e^2cos^2θ) \)