The area enclosed between the ellipse $9 x^{2} + 4 y^{2} - 36 x + 8 y + 4 = 0$ and the line $3 x + 2 y - 10 = 0$ in first Quadrant is

\frac{3 π}{2} - 1

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\frac{3 π}{2} - 2

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\frac{3 π}{2} - 3

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\frac{π}{3} - 4

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Solution

The correct option is C $\frac{3 π}{2} - 3$
$9 x^{2} + 4 y^{2} - 36 \times + 8 y + 4 = 0 \Rightarrow 9 (x - 2)^{2} + 4 (y + 1)^{2} = 36$
$\Rightarrow \frac{(x - 2)^{2}}{4} + \frac{(y + 1)^{2}}{9} = 1$
Let X = x-2; Y = y + 1 then ellipse equation is $\frac{X^{2}}{4} + \frac{Y^{2}}{9} = 1$
The line 3x + 2y -10 = 0 reduces to 3X + 2Y -6 =0
$∴$ Area= $\frac{1}{4} π .2 .3 - \frac{1}{2} .2 .3$
$= \frac{3 π}{2} - 3$

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