The area (in $sq.units$ ) bounded between curves $y^{2} - 4 y - 4 x + 12 = 0$ and $x^{2} - 4 x - 3 y + 10 = 0,$ is

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Solution

$C_{1} : y^{2} - 4 y - 4 x + 12 = 0$
can be writtern as
$(y - 2)^{2} = 4 (x - 2) \Rightarrow Y^{2} = 4 X$
and curve
$C_{2} : x^{2} - 4 x - 3 y + 10 = 0$ can be written as,
$(x - 2)^{2} = 3 (y - 2) \Rightarrow X^{2} = 3 Y$
$∵ (y - 2) = Y, (x - 2) = X$
By shifting origin to $(2, 2)$ bounded area does not change.
So, required area $= ∣ ∣ ∣ \frac{coff. of X \times coff. of Y}{3} ∣ ∣ ∣ = \frac{4 \times 3}{3} = 4 sq.units$

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