Find the (iii) coordinates of the foci, (iv) eccentricity \[\mathbf{16}{{\mathbf{x}}^{\mathbf{2}}}+\text{ }{{\mathbf{y}}^{\mathbf{2}}}=\text{ }\mathbf{16}\]

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

$\mathbf{16}{{\mathbf{x}}^{\mathbf{2}}}+\text{ }{{\mathbf{y}}^{\mathbf{2}}}=\text{ }\mathbf{16}$

CBSE | Class 11 | Ellipse | Maths | RS Aggarwal

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

$\mathbf{16}{{\mathbf{x}}^{\mathbf{2}}}+\text{ }{{\mathbf{y}}^{\mathbf{2}}}=\text{ }\mathbf{16}$

Given:

$\mathbf{16}{{\mathbf{x}}^{\mathbf{2}}}+\text{ }{{\mathbf{y}}^{\mathbf{2}}}=\text{ }\mathbf{16}$

Divide by

$16$

to both the sides, we get

$\frac{16}{16}{{x}^{2}}+\frac{1}{16}{{y}^{2}}=1$

$\frac{{{x}^{2}}}{1}+\frac{1}{16}{{y}^{2}}=1$

…(i)

Since,

$1\text{ }<\text{ }16$

So, above equation is of the form,

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

…(ii)

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

$\begin{array}{*{35}{l}} {{a}^{2}}=\text{ }16\text{ }and\text{ }{{b}^{2}}=\text{ }1 \\ \Rightarrow a\text{ }=\text{ }\surd 16\text{ }and\text{ }b\text{ }=\text{ }\surd 1 \\ \Rightarrow a\text{ }=\text{ }4\text{ }and\text{ }b\text{ }=\text{ }1 \\ \end{array}$

(iii) To find: Coordinates of the foci

We know that,

Coordinates of foci =

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