AP EAMCET Engineering 22 Apr 2019 Shift 1 Paper

Let the rectangle be inscribed in the ellipse,

x2

64

+

y2

16

=1
Let the coordinates be P(8‌cos‌θ,−4‌sin‌θ). Then, Q(−8‌cos‌θ,4‌sin‌θ),R(−8‌cos‌θ,4‌sin‌θ),S(8‌cos‌θ ,−4‌sin‌θ)
The area of rectangle is given by,
A=PQ×RS
=16‌cos‌θ×8‌sin‌θ
=64‌sin‌2‌θ
Differentiate w.r.t θ

dA

dθ

=128‌cos‌2‌θ

d2A

dθ2

=−256‌sin‌2‌θ
For the critical number of A,

dA

dθ

=0
4ab‌cos‌2‌θ=0
cos‌2‌θ=0
2θ=

π

2

or

3π

2

θ=

π

4

or

3π

4

Then,
(

d2A

dθ2

)θ=

π

4

=−256‌sin‌

π

2

=−256
<0
So, A is maximum when θ=

π

4

.
This implies,
PQ=l=16‌cos‌θ
=16‌cos‌

π

4

=16×

1

√2

=8√2
And,
PS=b=8‌sin‌θ
=8‌sin‌ ‌

π

4

=8×

1

√2

=4√2
Therefore, (l,b)=(8√2,4√2)