JEE MAIN - Mathematics (2021 - 16th March Evening Shift - No. 8)
Let C1 be the curve obtained by the solution of differential equation
$$2xy{{dy} \over {dx}} = {y^2} - {x^2},x > 0$$. Let the curve C2 be the
solution of $${{2xy} \over {{x^2} - {y^2}}} = {{dy} \over {dx}}$$. If both the curves pass through (1, 1), then the area enclosed by the curves C1 and C2 is equal to :
$$2xy{{dy} \over {dx}} = {y^2} - {x^2},x > 0$$. Let the curve C2 be the
solution of $${{2xy} \over {{x^2} - {y^2}}} = {{dy} \over {dx}}$$. If both the curves pass through (1, 1), then the area enclosed by the curves C1 and C2 is equal to :
$${\pi \over 4}$$ + 1
$$\pi$$ + 1
$$\pi$$ $$-$$ 1
$${\pi \over 2}$$ $$-$$ 1
Explanation
$${{dy} \over {dx}} = {{{y^2} - {x^2}} \over {2xy}}$$
Put $$y = vx$$
$$v + x{{dv} \over {dx}} = {{{v^2}{x^2} - {x^2}} \over {2v{x^2}}} = {{{v^2} - 1} \over {2v}}$$
$$x{{dv} \over {dx}} = {{{v^2} - 1 - 2{v^2}} \over {2v}} = - {{({v^2} + 1)} \over {2v}}$$
$$ \Rightarrow {{2v} \over {{v^2} + 1}}dv = - {{dx} \over x}$$
$$\ln ({v^2} + 1) = - \ln x + \ln c \Rightarrow {v^2} + 1 = {c \over x}$$
$$ \Rightarrow {{{y^2}} \over {{x^2}}} + 1 = {c \over x} \Rightarrow {x^2} + {y^2} = cx$$
It pass through (1, 1)
$$ \therefore $$ $${x^2} + {y^2} - 2x = 0$$
Similarly for second differential equation $${{dy} \over {dx}} = $$$${{2xy} \over {{x^2} - {y^2}}}$$
Equation of curve is x2 + y2 $$-$$ 2y = 0
Now required area is
_16th_March_Evening_Shift_en_8_1.png)
$$ = 2\int\limits_0^1 {\left( {\sqrt {2x - {x^2}} - x} \right)} dx $$
$$= ({\pi \over 2} - 1)$$ sq. units
Put $$y = vx$$
$$v + x{{dv} \over {dx}} = {{{v^2}{x^2} - {x^2}} \over {2v{x^2}}} = {{{v^2} - 1} \over {2v}}$$
$$x{{dv} \over {dx}} = {{{v^2} - 1 - 2{v^2}} \over {2v}} = - {{({v^2} + 1)} \over {2v}}$$
$$ \Rightarrow {{2v} \over {{v^2} + 1}}dv = - {{dx} \over x}$$
$$\ln ({v^2} + 1) = - \ln x + \ln c \Rightarrow {v^2} + 1 = {c \over x}$$
$$ \Rightarrow {{{y^2}} \over {{x^2}}} + 1 = {c \over x} \Rightarrow {x^2} + {y^2} = cx$$
It pass through (1, 1)
$$ \therefore $$ $${x^2} + {y^2} - 2x = 0$$
Similarly for second differential equation $${{dy} \over {dx}} = $$$${{2xy} \over {{x^2} - {y^2}}}$$
Equation of curve is x2 + y2 $$-$$ 2y = 0
Now required area is
$$ = 2\int\limits_0^1 {\left( {\sqrt {2x - {x^2}} - x} \right)} dx $$
$$= ({\pi \over 2} - 1)$$ sq. units
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