The differential equation $(3 x + 4 y + 1) d x + (4 x + 5 y + 1) d y = 0$ represents a family of

Circles

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Parabolas

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Ellipses

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Hyperbolas

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Solution

The correct option is B Hyperbolas
The given differential equation is
$(3 x + 4 y + 1) d x + (4 x + 5 y + 1) d y = 0... (i)$
Comparing eq. (i) with $M d x + N d y = 0$ ,
we get
$M = 3 x + 4 y + 1$
and $N = 4 x + 5 y + 1$
Here, $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} = 4$
Hence, eq. (i) is exact and solution is given by
$\int (3 x + 4 y + 1) d x + \int (4 x + 5 y + 1) d y = C$
$\Rightarrow \frac{3 x^{2}}{2} + 4 x y + x + 4 x y + \frac{5 y^{2}}{2} + y - C = 0$
$\Rightarrow 3 x^{2} + 16 x y + 2 x + 5 y^{2} + 2 y - 2 C = 0$
$\Rightarrow 3 x^{2} + 2 \times 8 x y + 2 x + 5 y^{2} + 2 y + C^{'} = 0 . . . (i i)$
where, $C^{'} = - 2 C$
On comparing eq. (ii) with standard form of conic section
$a x^{2} + 2 h x y + b y^{2} + 2 g x + 2 f y + C = 0$
We get,
$a = 3, h = 8, b = 5$
Here, $h^{2} - a b = 64 - 15 = 49 > 0$
Hence, the solution of differential equation represents family of hyperbolas.

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