QJEEM22Ma3eb3: JEE Main 2022 Mathematics - Class 12 JEEM22Ma3eb3 Solution: Consider a curve in the first quadrant as shown in the figure.…

${A_1} + {A_2} = xy - 8$ & ${A_1} = 2{A_2}$

${A_1} + {{{A_1}} \over 2} = xy - 8$

${A_1} = {2 \over 3}(xy - 8)$

$\int\limits_4^x {f(x)dx = {2 \over 3}(xf(x) - 8)}$

Differentiate w.r.t. x

$f(x) = {2 \over 3}\{ xf'(x) + f(x)\}$

${2 \over 3}xf'(x) = {1 \over 3}f(x)$

$2\int {{{f'(x)} \over {f(x)}}dx = \int {{{dx} \over x}} }$

$2\ln f(x) = \ln x + \ln c$

${f^2}(x) = cx$

Which passes through (4, 2)

$4 = c \times 4 \Rightarrow c = 1$

Equation of required curve

${y^2} = x$

Equation of normal having slope ( $-$ 6) is

$y = - 6x - 2\left( {{1 \over 4}} \right)( - 6) - {1 \over 4}{( - 6)^3}$

$y = - 6x + 57$

Which does not pass through $(10,\, - 4)$