Find the area bounded by $x^{2} + y^{2} + 4 x + 6 y - 3 = 0$ and $x^{2} + 4 x = 6 y + 14$ .

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Solution

$x^{2} - 4 x = 6 y + 14$
$x^{2} - 4 x + 4 = 6 y + 18$
$(x - 2)^{2} = 6 (y + 3)$
$x^{2} = 4 a y$
$4 a = 6$
$a = \frac{2}{3}$
Focus : $(a, 0)$
$x = a y = 0$
$x - 2 = \frac{2}{3} y + 3 = 0$
$x = 2 + \frac{2}{3} y = - 3$
$= \frac{8}{3}$
$(\frac{8}{3}, - 3)$

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Find the area bounded by x2+y2+4x+6y−3=0 and x2+4x=6y+14.

x2−4x=6y+14x2−4x+4=6y+18(x−2)2=6(y+3)x2=4ay4a=6a=23Focus : (a,0)x=ay=0x−2=23y+3=0x=2+23y=−3=83(83,−3)

Find the area bounded by $x^{2} + y^{2} + 4 x + 6 y - 3 = 0$ and $x^{2} + 4 x = 6 y + 14$ .

$x^{2} - 4 x = 6 y + 14$
$x^{2} - 4 x + 4 = 6 y + 18$
$(x - 2)^{2} = 6 (y + 3)$
$x^{2} = 4 a y$
$4 a = 6$
$a = \frac{2}{3}$
Focus : $(a, 0)$
$x = a y = 0$
$x - 2 = \frac{2}{3} y + 3 = 0$
$x = 2 + \frac{2}{3} y = - 3$
$= \frac{8}{3}$
$(\frac{8}{3}, - 3)$