Area Under Curves — JEE Main · 2024 · 01 Feb 2024 · Shift 2 Question with Solution

The sum of squares of all possible values of k , for which area of the region bounded by the parabolas

2 y^{2} = k x

and

k y^{2} = 2 (y - x)

is maximum, is equal to :

Answer: 8

Solution

From

2 y^{2} = k x

x = \frac{2 y ^{2}}{k}

. Substitute into

k y^{2} = 2 (y - x)

k y^{2} = 2 (y - \frac{2 y ^{2}}{k}) = 2 y - \frac{4 y ^{2}}{k} ⟹ y^{2} (k + \frac{4}{k}) = 2 y .

Besides

y = 0

(the origin), the curves meet at

y = \frac{2}{k + 4/ k} = \frac{2 k}{k ^{2} + 4}, x = \frac{2 y ^{2}}{k} = \frac{8 k}{( k ^{2} + 4 ) ^{2}} .

Writing

x = y - \frac{k y ^{2}}{2}

from the second curve, the enclosed area is

A = \int_{0}^{Y} [(y - \frac{k y ^{2}}{2}) - \frac{2 y ^{2}}{k}] d y = \int_{0}^{Y} [y - y^{2} (\frac{k}{2} + \frac{2}{k})] d y, Y = \frac{2 k}{k ^{2} + 4} .

Note

Y (\frac{k}{2} + \frac{2}{k}) = Y \cdot \frac{k ^{2} + 4}{2 k} = 1

, so

\frac{k}{2} + \frac{2}{k} = \frac{1}{Y}

. Then

A = \frac{Y ^{2}}{2} - \frac{Y ^{3}}{3} \cdot \frac{1}{Y} = \frac{Y ^{2}}{2} - \frac{Y ^{2}}{3} = \frac{Y ^{2}}{6} = \frac{1}{6} (\frac{2 k}{k ^{2} + 4})^{2} = \frac{2}{3} \cdot \frac{k ^{2}}{( k ^{2} + 4 ) ^{2}} .

Let

u = k^{2} > 0

A (u) = \frac{2}{3} \cdot \frac{u}{( u + 4 ) ^{2}}

\frac{d A}{d u} \propto \frac{( u + 4 ) ^{2} - u \cdot 2 ( u + 4 )}{( u + 4 ) ^{4}} = \frac{4 - u}{( u + 4 ) ^{3}} .

This vanishes at

u = 4

, i.e.

k^{2} = 4

k = \pm 2

— a maximum (the sign of

d A / d u

changes from positive to negative there).

Sum of squares of all possible values of

k

2^{2} + (- 2)^{2} = 4 + 4 = 8.

JEE Main · 2024 · 01 Feb 2024 · Shift 2 · Integer answer