The area of the greatest rectangle that can b | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Let a point on the ellipse be $(a \cos 	heta, b \sin 	heta)$.Since the rectangle is inscribed in the ellipse,its vertices are $(a \cos 	heta, b

Vedclass · Accepted Answer

$2ab$

Vedclass · Answer

(C) Let a point on the ellipse be $(a \cos 	heta, b \sin 	heta)$.Since the rectangle is inscribed in the ellipse,its vertices are $(a \cos 	heta, b \sin 	heta)$,$(-a \cos 	heta, b \sin 	heta)$,$(-a \cos 	heta, -b \sin 	heta)$,and $(a \cos 	heta, -b \sin 	heta)$.The length of the rectangle is $2a \cos 	heta$ and the breadth is $2b \sin 	heta$.Area of the rectangle $A = (2a \cos 	heta) 	imes (2b \sin 	heta) = 4ab \sin 	heta \cos 	heta = 2ab \sin 2	heta$.For the area to be the greatest,$\sin 2	heta$ must be maximum,i.e.,$\sin 2	heta = 1$.Therefore,the maximum area is $2ab(1) = 2ab$.

The area of the greatest rectangle that can be inscribed in the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is

Similar Questions

The value of $k$ such that the line $y=2x+k$ touches the ellipse $3x^2+5y^2=15$ is

If the eccentricity of an ellipse is $1/\sqrt{2}$,then its latus rectum is equal to its

Find the coordinates of the foci,the vertices,the length of the major axis,the minor axis,the eccentricity,and the length of the latus rectum of the ellipse $\frac{x^{2}}{25}+\frac{y^{2}}{100}=1$.

If a tangent with slope $-4/3$ to the ellipse $\frac{x^2}{18} + \frac{y^2}{32} = 1$ intersects the major axis and minor axis at $A$ and $B$ respectively,then the area of $\Delta OAB$ is .......... square units.

The sides of the rectangle of greatest area that can be inscribed in the ellipse $x^2+4y^2=64$ are :

Vedclass Test Series

Exam Paper Generator

Online Exam Module