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The equation $5x^2+9y^2+10x-36y-4=0$ represents

1. an ellipse with the coordinates of foci being $(\pm3,0)$
2. a hyperbola with the coordinates of foci being $(\pm3,0)$
3. an ellipse with the coordinates of foci being $(\pm2,0)$
4. a hyperbola with the coordinates of foci being $(\pm2,0)$
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It represents an ellipse

The general equation of conic section is represented as: $ax^{2}+ $$2 hx$$y$  + $b$$y^{2} + 2$$g$$x + 2$$f$$y + c = 0 or A$$x^{2}$ + $B$$x$$y$  + $C$$y^{2} + D$$x$ + $E$$y + F = 0 The determinant of this conic section is defined as : \Delta = \begin{vmatrix}a &h &g \\ h& b&f \\ g&f &c \end{vmatrix} = abc + 2fgh - a f^{2} - b g^{2} - c h^{2} (Refer this for the detailed explanations of the conic sections and the related conditions.) For the given equation, the value of \Delta < 0 and the value of B^{2} - 4AC < 0 and also A$$\neq$$C(again refer this for conditions) \Rightarrow the given equation represents an ellipse. Simplifying and rearranging the terms in the equation, 5$$x^{2}$ $+$ $10x$ $+$ $5$  $+$ $9$$y^{2} - 36y + 36 - 45 = 0 \Rightarrow 5$$\left ( x+2 \right )^{2}$  $+$ $9$$\left ( y-2 \right )^{2} = 45 \Rightarrow \left ( x+2 \right )^{2} / 9 + \left ( y-2 \right )^{2} / 5 = 1 \Rightarrow a^{2} = 9 and b^{2} = 5 Now, c = \pm$$\sqrt{a^{2}-b^{2}}$  $=$  $\pm$ $2$

$\therefore$ The ellipse is centered at  $\left ( -2,2 \right )$  and has focii at  $\left ( \pm2,0 \right )$

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