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

Three points

O (0, 0), P (a, a^{2}), Q (- b, b^{2}), a > 0, b > 0

, are on the parabola

y = x^{2}

. Let

S_{1}

be the area of the region bounded by the line PQ and the parabola, and

S_{2}

be the area of the triangle OPQ . If the minimum value of

\frac{S _{1}}{S _{2}}

\frac{m}{n}, gcd (m, n) = 1

, then

m + n

is equal to :

Answer: 7

Solution

O (0, 0)

P (a, a^{2})

Q (- b, b^{2})

, with

a, b > 0

, all on

y = x^{2}

.

Line

P Q

has slope

\frac{a ^{2} - b ^{2}}{a - ( - b )} = a - b

, so its equation is

y = (a - b) x + ab .

S_{1} = \int_{- b}^{a} [(a - b) x + ab - x^{2}] d x = (a - b) (a + b) (\frac{( a - b ) ^{2}}{2 ( a + b )}) \dots

Carrying out the integration directly:

S_{1} = [(a - b) \frac{x ^{2}}{2} + ab x - \frac{x ^{3}}{3}]_{- b}^{a} = \frac{( a - b ) ^{2} ( a + b )}{2} + ab (a + b) - \frac{a ^{3} + b ^{3}}{3} .

S_{2}

, the area of

△ O P Q

(with

O

at the origin):

S_{2} = \frac{1}{2} x_{P} y_{Q} - x_{Q} y_{P} = \frac{1}{2} a b^{2} - (- b) (a^{2}) = \frac{1}{2} ab (a + b) .

Dividing and simplifying the algebra:

\frac{S _{1}}{S _{2}} = \frac{1}{3} (\frac{a}{b} + \frac{b}{a} + 2) .

By AM–GM,

\frac{a}{b} + \frac{b}{a} \geq 2

, with equality when

a = b

. So the minimum value of

\frac{S _{1}}{S _{2}}

\frac{1}{3} (2 + 2) = \frac{4}{3} .

With

\frac{m}{n} = \frac{4}{3}

g cd (4, 3) = 1

, so

m + n = 4 + 3 = 7

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