A man running a racecourse notes that the sum of the distances from the two flag posts from him is always 10 m and the distance between the flag posts is 8 m. Find the equation of the posts traced by the man
A man running a racecourse notes that the sum of the distances from the two flag posts from him is always 10 m and the distance between the flag posts is 8 m. Find the equation of the posts traced by the man

Leave

    \[A\text{ }and\text{ }B\]

alone the places of the two banner posts and

    \[P\left( x,\text{ }y \right)\]

be the situation of the man.

In this way,

    \[PA\text{ }+\text{ }PB\text{ }=\text{ }10.\]

We realize that if a point moves in plane so that the amount of its separation from two fixed point is steady, then, at that point, the way is a circle and this consistent worth is equivalent to the length of the significant hub of the oval.

Then, at that point, the way depicted by the man is a circle where the length of the significant hub is

    \[10m\]

, while focuses

    \[A\text{ }and\text{ }B\]

are the foci.

Presently let us take the beginning of the facilitate plane as the focal point of the circle, and taking the significant hub along the

    \[x-pivot,\]

The diagrammatic portrayal of the oval is as per the following:

NCERT Solutions for Class 11 Maths Chapter 11 – Conic Sections image - 7

The condition of the oval is as

    \[{{x}^{2}}/{{a}^{2}}~+\text{ }{{y}^{2}}/{{b}^{2}}~=\text{ }1\]

, where ‘

    \[a\]

‘ is the semi-significant hub.

Thus,

    \[2a\text{ }=\text{ }10\]

    \[\begin{array}{*{35}{l}} a\text{ }=\text{ }10/2  \\ =\text{ }5  \\ \end{array}\]

Distance between the foci

    \[2c\text{ }=\text{ }8\]

    \[\begin{array}{*{35}{l}} c\text{ }=\text{ }8/2  \\ =\text{ }4  \\ \end{array}\]

By utilizing the connection,

    \[c\text{ }=~\surd ({{a}^{2}}~\text{ }{{b}^{2}})\]

,we get

    \[<span class="ql-right-eqno"> (1) </span><span class="ql-left-eqno">   </span><img src="https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-457ca8b04f343b7a18bcb9d57531a779_l3.png" height="198" width="215" class="ql-img-displayed-equation quicklatex-auto-format" alt="\begin{align*} & 4\text{ }=~\surd \left( 25\text{ }\text{ }{{b}^{2}} \right) \\ & \begin{array}{*{35}{l}} 16\text{ }=\text{ }25\text{ }\text{ }{{b}^{2}}  \\ {{b}^{2}}~=\text{ }25\text{ }-1  \\ =\text{ }9  \\ b\text{ }=\text{ }3  \\ \end{array} \\ \end{align*}" title="Rendered by QuickLaTeX.com"/>\]

Henceforth, condition of the way followed by the man

    \[{{x}^{2}}/25\text{ }+\text{ }{{y}^{2}}/9\text{ }=\text{ }1\]