Login/Sign Up
Home » Find the (iii) coordinates of the foci, (iv) eccentricity    
Find the (iii) coordinates of the foci, (iv) eccentricity

    \[\frac{{{x}^{2}}}{4}+\frac{{{y}^{2}}}{25}=1\]

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

    \[\frac{{{x}^{2}}}{4}+\frac{{{y}^{2}}}{25}=1\]

Given:

    \[\frac{{{x}^{2}}}{4}+\frac{{{y}^{2}}}{25}=1\]

…(i)

Since,

    \[4\text{ }<\text{ }25\]

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

    \[\begin{array}{*{35}{l}}</strong> <strong> {{a}^{2}}=\text{ }25\text{ }and\text{ }{{b}^{2}}=\text{ }4 \\</strong> <strong> \Rightarrow a\text{ }=\text{ }\surd 25\text{ }and\text{ }b\text{ }=\text{ }\surd 4 \\</strong> <strong> \Rightarrow a\text{ }=\text{ }5\text{ }and\text{ }b\text{ }=\text{ }2 \\</strong> <strong>\end{array}\]

(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

    \[<span class="ql-right-eqno"> (1) </span><span class="ql-left-eqno"> </span><img src="https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-b9968e0334e645ab0158e40d401f0a92_l3.png" height="106" width="147" class="ql-img-displayed-equation quicklatex-auto-format" alt="\begin{align*} & \begin{array}{*{35}{l}} {{c}^{2}}=\text{ }{{a}^{2}}\text{ }{{b}^{2}} \\ =\text{ }25\text{ }\text{ }4 \\ {{c}^{2}}=\text{ }21 \\ \end{array} \\ & \\ & \\ \end{align*}" title="Rendered by QuickLaTeX.com"/>\]

    \[c\text{ }=\text{ }\surd 21\]

…(I)

∴ Coordinates of foci =

    \[\left( 0,\text{ }\pm \surd 21 \right)\]

(iv) To find: Eccentricity

We know that,

Eccentricity e=c/a

    \[e=\frac{\sqrt{21}}{5}\]

[from(I)]

i

More Material