The area of the rectangle formed by the perpendiculars from the centre of the standard ellipse to the tangent and normal at its point whose eccentric angles `pi/4` is

The area of the rectangle formed by the perpendiculars from the centre of the ellipse (x^(2))/(9)+(y^(2))/(4)=1 to the tangent and normal at the point whose eccentric angle is (pi)/(4) , is :

The area of the rectangle formed by the perpendicular from centre of (x^(2))/(9)+(y^(2))/(4)=1 to the tangents area normal the point where occeses angle is (pi)/(4) equals

If area of rectangle formed by the perpendiculars drawn from centre of ellipse (x^(2))/(9)+(y^(2))/(16)=1 to the tangent and normal at the point (P((3)/(sqrt(2)),2sqrt(2))) on ellipse is ((a)/(b)) (where a and b are coprime numbers) then (a-3b) equals

The equation of the normal to the ellipse at the point whose eccentric angle theta=(pi)/(6) is

Find the locus of the foot of perpendicular from the centre of the ellipse (x^(2))/(a^(2))+(y^(2))/(b^(2))=1 on the chord joining the points whose eccentric angles differ by (pi)/(2)

In an ellipse,the ratio of the area of the rectangle formed by the end points of its each latus rectum to that of the ellipse is (1)/(pi) .The eccentricity of the ellipse is

The equation of the normal to the ellipse x^(2)//16+y^(2)//9=1 at the point whose eccentric angle theta=pi//6 is