दिखाइए कि `(sqrt(a^(2)-b^(2)),0)` और `(-sqrt(a^(2)-b^(2)),0 )` बिन्दुओ से रेखा प`(x )/(a)cos theta + (y)/(b) sin theta =1` पर खींचे गये लम्बो कि लंबाइयाँ का गुणनफल `b ^(2 )` है ।

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)cos theta+(y)/(b)sin theta=1 is b^(2)

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 cos theta + y/b sin theta = 1 " is " b^(2) .

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 cos theta + y/b sin theta = 1 " is " b^(2) .

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 cos theta + y/b sin theta = 1 " is " b^(2) .

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) cos theta + (y)/( b) sin theta =1 is b^2 .

If x/a cos theta + y/b sin theta = 1 , x/a sin theta - y/b cos theta = 1 then x^2/a^2 + y^2/b^2 =

The product of the perpendiculars drawn from the points pm sqrt(a^(2) -b^(2),0) on the line x/a cos theta+ y/b sin theta =1, is

(x) / (a) cos theta + (y) / (b) sin theta = 1, (x) / (a) sin theta- (y) / (b) cos theta = 1 then (x ^ (2)) / (a ^ (2)) + (y ^ (2)) / (b ^ (2)) =

If x gt 0, a = x cos theta , and b = x sin theta , then sqrt(a^(2)+b^(2))=