Question

Coordinates of the foci of the ellipse $5x^2+9y^2+10x-36y-4=0,$ are

• A. $\left(1,2\right)\mathrm{and}\left(-3,2\right)$
• B. $\left(2,1\right)\mathrm{and}\left(-3,2\right)$
• C. $\left(1,2\right)\mathrm{and}\left(3,2\right)$
• D. none of these

Answer 1

The correct option is A

$\left(1,2\right)\mathrm{and}\left(-3,2\right)$

Explanation for correct option:

Step-1: Reduce into standard form.

Given, equation $5x^2+9y^2+10x-36y-4=0$

$\Rightarrow$${\left(\sqrt{5}x\right)}^2+{\left(3y\right)}^2+2\times \sqrt{5}\times \sqrt{5}\times x-2\times 3\times 6\times y+{\left(6\right)}^2+{\left(\sqrt{5}\right)}^2-45=0$

$\Rightarrow$ ${\left(\sqrt{5}x+\sqrt{5}\right)}^2+{\left(3y-6\right)}^2=45$

$\Rightarrow$ $\frac{{\left(x+1\right)}^2}{{\left(3\right)}^2}+\frac{{\left(y+\left(-2\right)\right)}^2}{{\left(\sqrt{5}\right)}^2}=1........\left(i\right)$

Step-2: Compare with standard form.

Equation of an ellipse in standard form is $\frac{{\left(x-x_1\right)}^2}{a^2}+\frac{{\left(y-y_1\right)}^2}{b^2}=1.......\left(ii\right)$

Compare equation $\left(i\right)$ and $\left(ii\right)$

$$\begin{gathered} x_1=-1 \\ {y}_1=2 \\ a=3 \\ b=\sqrt{5} \end{gathered}$$

Step-3: Finding foci.

We know that foci is $\left(x_1\pm c,c\right)$

$c=\sqrt{a^2-b^2}$

$\Rightarrow$ $c=\sqrt{9-5}$

$\Rightarrow$ $c=\pm 2$

Thus, foci is $\left(1,2\right)\&\left(-3,2\right)$

Hence, correct answer is option $\left(A\right)$.