Prove that the product of the lengths of the perpendiculars drawn from the points `(sqrt(a^2-b^2),0)`and `(-sqrt(a^2-b^2),0)`to the line `x/a``costheta``+``y/b``sintheta=1` is `b^2`.

Explanation

`p=` length of perpendicular from point -`(sqrt(a^(2)-b^(2)),0)` to the line `(xcostheta)/(a)+(ysintheta)/(b)=1`
`impliesp=|((sqrt(a^(2)-b^(2)))/(a)*costheta+0-1)/(sqrt((cos^(2)theta)/(a^(2))+(sin^(2)theta)/(b^(2))))|`
`=|(bsqrt(a^(2)-b^(2)*costheta-a))/sqrt(b^(2)cos^(2)theta+a^(2)sin^(2)theta)|`
and `q=` length of perpendicular from point `(-sqrt(a^(2)-b^(2)),0)` to the line `(xcostheta)/(a)+(ysintheta)/(b)=1`
`impliesq=|((-sqrt(a^(2)-b^(2))*costheta)/(a)+0-1)/(sqrt(cos^(2)theta)/(a^(2))+(sin^(2)theta)/(b^(2)))|`
`=|(b(-sqrt(a^(2)-b^(2))*costheta-a))/(sqrt(b^(2)cos^(2)theta+a^(2)sin^(2)theta))|`
Now`p*q=|(b(sqrt(a^(2)-b^(2))*costheta-a))/(sqrt(b^(2)cos^(2)theta+a^(2)sin^(2)theta))|*|(b(-sqrt(a^(2)-b^(2))*costheta-a))/(sqrt(b^(2)cos^(2)theta+a^(2)sin^(2)theta))|`
`=|(b^(2){a^(2)-(a^(2)-b^(2))cos^(2)theta})/(b^(2)cos^(2)theta+a^(2)sin^(2)theta)|`
`=b^(2)|(a^(2)-a^(2)cos^(2)theta+b^(2)cos^(2)theta)/(b^(2)cos^(2)theta+a^(2)sin^(2)theta)|`
`=b^(2)|(a^(2)(1-cos^(2)0)+b^(2)cos^(2)theta)/(b^(2)cos^(2)theta+a^(2)sin^(2)theta)|`
`b^(2)|(a^(2)sin^(2)theta+b^(2)cos^(2)theta)/(b^(2)cos^(2)theta+a^(2)sin^(2)theta)|=b^(2)`