The area enclosed by the closed curve C given by the differential equation d y d x + x + a y - 2 = 0 , y ( 1 ) = 0 is 4. Let P and Q be the points of intersection of the curve C and the y -axis. If normals at P and Q on the curve C intersect the x

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Solution

Q.
The area enclosed by the closed curve C given by the differential equation $\frac{d y}{d x} + \frac{x + a}{y - 2} = 0, y (1) = 0$ is $4 π$ . Let P and Q be the points of intersection of the curve C and the $y$ -axis. If normals at P and Q on the curve C intersect the $x$ -axis at points R and S respectively, then the length of the line segment RS is [2023]

1 $\frac{4 \sqrt{3}}{3}$

2 $\frac{2 \sqrt{3}}{3}$

3 $2$

4 $2 \sqrt{3}$

Ans.
(1)

$\frac{d y}{d x} + \frac{x + a}{y - 2} = 0, y (1) = 0;$ $\frac{d y}{d x} = \frac{- (x + a)}{y - 2}$

$\Rightarrow \int (y - 2) d y = - \int (x + a) d x \Rightarrow \frac{y^{2}}{2} - 2 y = \frac{- x^{2}}{2} - a x + c$

$\Rightarrow \frac{x^{2} + y^{2}}{2} + a x - 2 y = c \Rightarrow x^{2} + y^{2} + 2 a x - 4 y = c_{1} (∵ c_{1} = 2 c)$

$Substituting x = 1 and y = 0 as y (1) = 0$

$\Rightarrow 1 + 2 a = c_{1} ∴ x^{2} + y^{2} + 2 a x - 4 y = 1 + 2 a$

$This represents the equation of a circle having area 4 π$

$∴$ $Radius of circle = 2$

$Now r^{2} = 2^{2} = {(- a)}^{2} + {(2)}^{2} = 1 + 2 a$

$\Rightarrow 0 = {(a + 1)}^{2} \Rightarrow a = - 1$

$∴$ $Equation of circle is x^{2} + y^{2} - 2 x - 4 y = - 1$

$\Rightarrow x^{2} - 2 x + 1 + y^{2} - 4 y + 4 - 4 = 0$

$\Rightarrow {(x - 1)}^{2} + {(y - 2)}^{2} = 4$

$Now, substitute x = 0 to get intersection point of circle with y-axis$

$\Rightarrow {(y - 2)}^{2} = 3$

$\Rightarrow y = \pm \sqrt{3} + 2$ $∴$ $P (0, \pm \sqrt{3} + 2)$

$P (0, \pm \sqrt{3} + 2) and Q (0, 2 - \sqrt{3}) are the points.$

$Now normal at P will pass through centre (1, 2)$

$∴$ $Equation of line passing through P and (1, 2) is$

$y - (\sqrt{3} + 2) = \frac{- \sqrt{3}}{1} (x)$

$Now, this line will cut x-axis at R (\frac{\sqrt{3} + 2}{\sqrt{3}}, 0)$

$Also, equation of line passing through Q (0, 2 - \sqrt{3}) and (1, 2) is$ $y - (2 - \sqrt{3}) = \sqrt{3} x$

$This will cut the x-axis at S (\frac{\sqrt{3} - 2}{\sqrt{3}}, 0)$

$R S = \sqrt{{(\frac{\sqrt{3} + 2}{\sqrt{3}} - \frac{\sqrt{3} - 2}{\sqrt{3}})}^{2} + {(0 - 0)}^{2}} = \sqrt{{(\frac{4}{\sqrt{3}})}^{2}} = \frac{4}{\sqrt{3}} = \frac{4 \sqrt{3}}{3}$

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