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

- an ellipse with the coordinates of foci being $(\pm3,0)$
- a hyperbola with the coordinates of foci being $(\pm3,0)$
- an ellipse with the coordinates of foci being $(\pm2,0)$
- a hyperbola with the coordinates of foci being $(\pm2,0)$

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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 $\left ( \pm2,0 \right )$

Option C is the answer.

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