Find the equation of the ellipse with center at the origin, the major axis on the x-axis and passing through the points

    \[\left( \mathbf{4},\text{ }\mathbf{3} \right)\]

and

    \[(-1,4)\]

.
Find the equation of the ellipse with center at the origin, the major axis on the x-axis and passing through the points

    \[\left( \mathbf{4},\text{ }\mathbf{3} \right)\]

and

    \[(-1,4)\]

.

Given:

Center is at the origin

and Major axis is along x – axis

So, Equation of ellipse is of the form

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

…(i)

Given that ellipse passing through the points

    \[\left( \mathbf{4},\text{ }\mathbf{3} \right)\]

and

    \[(-1,4)\]

So, point

    \[\left( \mathbf{4},\text{ }\mathbf{3} \right)\]

and

    \[(-1,4)\]

will satisfy the eq. (i)

Taking point

    \[\left( \mathbf{4},\text{ }\mathbf{3} \right)\]

where x =

    \[4\]

and y =

    \[3\]

Putting the values in eq. (i), we get

    \[\frac{{{(4)}^{2}}}{{{a}^{2}}}+\frac{{{(3)}^{2}}}{{{b}^{2}}}=1\]

    \[\frac{16}{{{a}^{2}}}+\frac{9}{{{b}^{2}}}=1\]

…(ii)

Taking point

    \[(-1,4)\]

where x =

    \[-1\]

and y =

    \[4\]

Putting the values in eq. (i), we get

    \[\frac{16}{{{a}^{2}}}+\frac{16\times 16}{{{b}^{2}}}=16\]

    \[\frac{16}{{{a}^{2}}}+\frac{256}{{{b}^{2}}}=16\]

…(iii)

Now, we have to solve the above two equations to find the value of a and b Multiply the eq. (iii) by 16, we get

    \[\frac{16}{{{a}^{2}}}+\frac{16\times 16}{{{b}^{2}}}=16\]

 

    \[\frac{16}{{{a}^{2}}}+\frac{256}{{{b}^{2}}}=16\]

…(iv)

Subtracting eq. (iv) from (ii), we get

    \[<span class="ql-right-eqno"> (1) </span><span class="ql-left-eqno">   </span><img src="https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-2d8d38e21d8df273271b01526fdafeed_l3.png" height="215" width="206" class="ql-img-displayed-equation quicklatex-auto-format" alt="\begin{align*} & \frac{9-256}{{{b}^{2}}}=-15 \\ & -\frac{247}{{{b}^{2}}}=-15 \\ & {{b}^{2}}=\frac{247}{15} \\ \end{align*}" title="Rendered by QuickLaTeX.com"/>\]

Substituting the value of

    \[{{b}^{2}}\]

in eq. (iii), we get

TThus,

    \[{{a}^{2}}=\frac{247}{7}\,\,\,\And \,\,{{b}^{2}}=\frac{247}{15}\]

Substituting the value of

    \[{{a}^{2}}and\text{ }{{b}^{2}}\]

in eq. (i), we get

    \[<span class="ql-right-eqno"> (2) </span><span class="ql-left-eqno">   </span><img src="https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-03e3a2b095eebf414e0cb221ae9cd61b_l3.png" height="261" width="570" class="ql-img-displayed-equation quicklatex-auto-format" alt="\begin{align*}</strong> <strong>  & \frac{{{x}^{2}}}{\frac{247}{7}}+\frac{{{y}^{2}}}{\frac{247}{15}}=1 \\ </strong> <strong> & \frac{7{{x}^{2}}}{247}+\frac{15{{y}^{2}}}{247}=1 \\ </strong> <strong> & 7{{x}^{2}}+15{{y}^{2}}=247 \\ </strong> <strong>\end{align*}" title="Rendered by QuickLaTeX.com"/>\]