A rectangle of maximum area is inscribed in a | vedclass

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

Question

(C) Let the point $P$ on the ellipse be $(5 \cos 	heta, 4 \sin 	heta)$ in the first quadrant.Since the rectangle is symmetric about both axes,its ve

Vedclass · Accepted Answer

$4 \sqrt{2}, 5 \sqrt{2}$

Vedclass · Answer

(C) Let the point $P$ on the ellipse be $(5 \cos 	heta, 4 \sin 	heta)$ in the first quadrant.Since the rectangle is symmetric about both axes,its vertices are $(\pm 5 \cos 	heta, \pm 4 \sin 	heta)$.The length of the rectangle is $L = 2(5 \cos 	heta) = 10 \cos 	heta$.The breadth of the rectangle is $B = 2(4 \sin 	heta) = 8 \sin 	heta$.The area $A$ of the rectangle is $A = L 	imes B = (10 \cos 	heta)(8 \sin 	heta) = 80 \sin 	heta \cos 	heta = 40 \sin(2 	heta)$.For the area to be maximum,$\sin(2 	heta)$ must be maximum,i.e.,$\sin(2 	heta) = 1$.This implies $2 	heta = \frac{\pi}{2}$,so $	heta = \frac{\pi}{4}$.Substituting $	heta = \frac{\pi}{4}$ into the expressions for length and breadth:$L = 10 \cos(\frac{\pi}{4}) = 10 	imes \frac{1}{\sqrt{2}} = 5 \sqrt{2}$.$B = 8 \sin(\frac{\pi}{4}) = 8 	imes \frac{1}{\sqrt{2}} = 4 \sqrt{2}$.Thus,the dimensions of the rectangle are $5 \sqrt{2}$ and $4 \sqrt{2}$.

$A$ rectangle of maximum area is inscribed in an ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1$. Then its dimensions are:

Similar Questions

$P$ is a point on the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. When the area of $\Delta PSS'$ is maximum,the inradius of $\Delta PSS'$ ($S$ and $S'$ are foci) is equal to:

If $S$ and $S^{\prime}$ are the foci of the ellipse $\frac{x^2}{18}+\frac{y^2}{9}=1$ and $P$ is a point on the ellipse,then $\min \left(SP \cdot S^{\prime}P\right) + \max \left(SP \cdot S^{\prime}P\right)$ is equal to:

Find the equation of the ellipse whose center is at $(2, -3)$,focus is at $(3, -3)$,and one vertex is at $(4, -3)$.

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

The focal distances of the point $\left(\frac{4}{\sqrt{5}}, \frac{3}{\sqrt{5}}\right)$ on the ellipse $\frac{x^2}{4}+\frac{y^2}{9}=1$ are

Vedclass Test Series

Exam Paper Generator

Online Exam Module