Find the (iii) coordinates of the foci, (iv) eccentricity

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Find the (iii) coordinates of the foci, (iv) eccentricity

Given:

$\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1$

…(i)

Since,

$25>9$

So, above equation is of the form,

$\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1$

…(ii)

Comparing eq. (i) and (ii), we get

$\begin{array}{*{35}{l}} {{a}^{2}}=\text{ }25\text{ }and\text{ }{{b}^{2}}=\text{ }9 \\ \Rightarrow a\text{ }=\text{ }\surd 25\text{ }and\text{ }b\text{ }=\text{ }\surd 9 \\ \Rightarrow a\text{ }=\text{ }5\text{ }and\text{ }b\text{ }=\text{ }3 \\ \end{array}$

(iii) To find: Coordinates of the foci

We know that,

Coordinates of foci =

$\left( \pm c,\text{ }0 \right)$

where

${{c}^{2}}=\text{ }{{a}^{2}}\text{ }{{b}^{2}}$

So, firstly we find the value of c

$\begin{array}{*{35}{l}} {{c}^{2}}=\text{ }{{a}^{2}}\text{ }{{b}^{2}} \\ =\text{ }25\text{ }\text{ }9 \\ {{c}^{2}}=\text{ }16 \\ c\text{ }=\text{ }\surd 16 \\ \end{array}$