Home
Class 12
MATHS
Area of the parallelogram formed by the ...

Area of the parallelogram formed by the tangents at the points whose eccentric angles are `theta,theta+(pi/2),theta+pi,theta+((3pi)/2)` on the ellipse `16x^2 + 25y^2 = 400` is

Promotional Banner

Similar Questions

Explore conceptually related problems

The area of the parallelogram formed by the tangents at the points whose eccentric angles are theta, theta +(pi)/(2), theta +pi, theta +(3pi)/(2) on the ellipse (x^(2))/(a^(2))+(y^(2))/(b^(2)) =1 is

The area of the parallelogram formed by the tangents at the points whose eccentric angles are theta, theta +(pi)/(2), theta +pi, theta +(3pi)/(2) on the ellipse (x^(2))/(a^(2))+(y^(2))/(b^(2)) =1 is

The area of the parallelogram formed by the tangents at the points whose eccentric angles are theta, theta +(pi)/(2), theta +pi, theta +(3pi)/(2) on the ellipse (x^(2))/(a^(2))+(y^(2))/(b^(2)) =1 is

The area of the parallelogram formed by the tangents at the points whose eccentric angles are theta, theta +(pi)/(2), theta +pi, theta +(3pi)/(2) on the ellipse (x^(2))/(a^(2))+(y^(2))/(b^(2)) =1 is

The area of the parallelogram formed by the tangents at the points whose eccentric angles are theta, theta + pi/2, theta + pi, theta +(3pi)/2 on the ellipse (x^(2))/(a^(2))+(y^(2))/(b^(2))=1 is

Locus of the point of intersection of the tangents at the points with eccentric angle theta and (pi)/(2)+theta is

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

If pi < theta<(3pi)/2' then theta lies in

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