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Find the (iii) coordinates of the foci, (iv) eccentricity
CBSE | Class 11 | Ellipse | Maths | RS Aggarwal
Find the (iii) coordinates of the foci, (iv) eccentricity

Given:

    \[\mathbf{9}{{\mathbf{x}}^{\mathbf{2}}}+\text{ }\mathbf{16}{{\mathbf{y}}^{\mathbf{2}}}=\text{ }\mathbf{144}\]

Divide by

    \[144\]

to both the sides, we get

    \[\frac{9}{144}{{x}^{2}}+\frac{16}{144}{{y}^{2}}=1\]

    \[\frac{{{x}^{2}}}{16}+\frac{{{y}^{2}}}{9}=1\]

…(i)

Since, 

    \[16>9\]

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

    \[\begin{array}{*{35}{l}} {{a}^{2}}=\text{ }16\text{ }and\text{ }{{b}^{2}}=\text{ }9 \\ \Rightarrow a\text{ }=\text{ }\surd 16\text{ }and\text{ }b\text{ }=\text{ }\surd 9 \\ \Rightarrow a\text{ }=\text{ }4\text{ }and\text{ }b\text{ }=\text{ }3 \\ \end{array}\]

(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

    \[\begin{array}{*{35}{l}} {{c}^{2}}=\text{ }{{a}^{2}}\text{ }{{b}^{2}} \\ =\text{ }16\text{ }\text{ }9 \\ {{c}^{2}}=\text{ }7 \\ \end{array}\]

    \[c\text{ }=\text{ }\surd 7\]

…(I)

∴ Coordinates of foci =

    \[\left( \pm \surd 7,\text{ }0 \right)\]

(iv) To find: Eccentricity

/ We know that,

Eccenctricity = c/a

    \[e=\frac{\sqrt{7}}{4}\]

   [from(I)]

i

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