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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

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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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