Find the (v) length of the latus rectum of each of the following ellipses.

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

Find the (v) length of the latus rectum of each of the following ellipses.

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

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

    \[\begin{array}{*{35}{l}}</strong> <strong>   {{a}^{2}}=\text{ }16\text{ }and\text{ }{{b}^{2}}=\text{ }1  \\</strong> <strong>   \Rightarrow a\text{ }=\text{ }\surd 16\text{ }and\text{ }b\text{ }=\text{ }\surd 1  \\</strong> <strong>   \Rightarrow a\text{ }=\text{ }4\text{ }and\text{ }b\text{ }=\text{ }1  \\</strong> <strong>\end{array}\]

(v) To find: Length of the Latus Rectum

We know that,

Length of the Latus Rectum =

    \[\frac{2{{b}^{2}}}{a}\]

    \[<span class="ql-right-eqno"> (1) </span><span class="ql-left-eqno">   </span><img src="https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-a1aca6f8213870eb5ca28128b5ea11a4_l3.png" height="224" width="107" class="ql-img-displayed-equation quicklatex-auto-format" alt="\begin{align*} & =\frac{2{{(1)}^{2}}}{4} \\ & =\frac{2}{4} \\ & =\frac{1}{2} \\ \end{align*}" title="Rendered by QuickLaTeX.com"/>\]