JEE MAIN - Mathematics (2022 - 27th July Evening Shift - No. 9)
Consider a curve $$y=y(x)$$ in the first quadrant as shown in the figure. Let the area $$\mathrm{A}_{1}$$ is twice the area $$\mathrm{A}_{2}$$. Then the normal to the curve perpendicular to the line $$2 x-12 y=15$$ does NOT pass through the point.
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Explanation
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$${A_1} + {A_2} = xy - 8$$ & $${A_1} = 2{A_2}$$
$${A_1} + {{{A_1}} \over 2} = xy - 8$$
$${A_1} = {2 \over 3}(xy - 8)$$
$$\int\limits_4^x {f(x)dx = {2 \over 3}(xf(x) - 8)} $$
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 $$(10,\, - 4)$$
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