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 $r^2=b^2/(1-e^2cos^2θ)$