KVPY SX SB Exam 19-Nov-2017 Question Paper

Question : 81 of 120

Marks: +1, -0

part - II

Mathematics

Let AB be the latus rectum of the parabola

y^2 = 4ax

in the xy-plane. Let T be the region bounded by the finite arc AB of the parabola and the line segment AB. A rectangle PQRS of maximum possible area is inscribed in T with P, Q on line AB, and R, S on arc AB. Then area(PQRS)/area(T) equals

$1/2$
$1/3$
$1/√2$
$1/√3$

Validate

Solution:

T = ∫↖{a}↙{0}2√a\ √x.dx={8a^2}/3

t_1 = t_2

Area of PQRS

= 2a | t_1-t_-2| × | a -at^2_1 |

But

t_1 = - t_2

= 4a^2 t_1 (1 -t^2_1 )

Differentiation with respect

t_1

We will get

t^2_1=1/3

Now put

t_1 =1/√3

get Area of
PQRS =

{8a^2}/{3√3}

ratio Becomes

1/√3

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