Find the area of the figure enclosed by the curve `5x^2+6x y+2y^2+7x+6y+6=0.`

Explanation

`5x^(2)+6xy+2y^(2)+7x+6y+6=0`
`2y^(2)+6(1+x)y+5x^(2)+7x+6=0 rArr y = (-3(1+x)+-sqrt((3-x)(x-1)))/(2)`
Clearly the vlues of y are real for `x[1,3]`

when `x = 1`, we get `y = - 3` and , `x = 3 rArr = - 6`
Let `f(x) = (-3(1+x)+sqrt((3-x)(x-1)))/(2)` and, `g(x) = (-3(1+x)-sqrt((3-x))(x-1))/(2)`
required area `= |underset(1)overset(3)int{g(x)-f(x)}dx|`
`=|underset(1)overset(3)intsqrt(-x^(2)+4x-3)|dx= |underset(1)overset(3)intsqrt(1^(2)-(x-2)^(3))|dx=|[1/2(x-2)sqrt(-x^(2)+4x-3)+1/2sin^(-1)((x-2)/(1))]_(1)^(3)|`
`= |[{1/2sin^(-1)}-{1/2sin^(-1)(-1)}]|=pi/2` sq. unit