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Hint:Draw a rough diagram and try to find the lengths of the semi-major axis and the semi-minor axis of the ellipse. After you find them use the formula of eccentricity to get the answer.
Complete step-by-step answer:
Starting by drawing a rough diagram:
Two points inside an ellipse that are used in its formal definition.The foci always lie on the major (longest) axis, spaced equally each side of the center. If the major axis and minor axis are the same length, the figure is a circle and both foci are at the center.
Now let us try to relate the diagram with the question.
From the figure, it is clear that the major axis of the ellipse is the x-axis, and the minor axis is the y-axis. Hence, the meeting point of the axes is O(0,0), which becomes the centre of the ellipse.
As the focus is given by (ae,0) and (-ae,0), we get
ae=4
Now we will put the value of e as given in the question. On doing so, we get
$\begin{align}
& a\times \dfrac{1}{3}=4 \\
& \Rightarrow a=12 \\
\end{align}$
Now the eccentricity of an ellipse is given by the formula: $e=\sqrt{1-\dfrac{{{\left( \text{length of semi-minor axis} \right)}^{2}}}{{{\left( \text{length of semi-major axis} \right)}^{2}}}}$ . Using this formula, we get
\[e=\sqrt{1-\dfrac{{{b}^{2}}}{{{12}^{2}}}}\]
Now using the value of e given in the question.
\[\dfrac{1}{3}=\sqrt{1-\dfrac{{{b}^{2}}}{{{12}^{2}}}}\]
\[\Rightarrow \dfrac{1}{9}=1-\dfrac{{{b}^{2}}}{{{12}^{2}}}\]
\[\Rightarrow b=\sqrt{\dfrac{8}{9}\times 144}=8\sqrt{2}\]
Therefore, the equation of the ellipse is:
$\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$
$\Rightarrow \dfrac{{{x}^{2}}}{144}+\dfrac{{{y}^{2}}}{128}=1$
Note: In mathematics, an ellipse is a plane curve surrounding two focal points, such that for all points on the curve, the sum of the two distances to the focal points is a constant.For these types of questions diagram is key point to approach the solution and students should know the definitions and parameters of ellipse i.e foci,vertices,major and minor axes and its equations ,eccentricity and latus rectum to solve these types of questions.
Complete step-by-step answer:
Starting by drawing a rough diagram:
Two points inside an ellipse that are used in its formal definition.The foci always lie on the major (longest) axis, spaced equally each side of the center. If the major axis and minor axis are the same length, the figure is a circle and both foci are at the center.
Now let us try to relate the diagram with the question.
From the figure, it is clear that the major axis of the ellipse is the x-axis, and the minor axis is the y-axis. Hence, the meeting point of the axes is O(0,0), which becomes the centre of the ellipse.
As the focus is given by (ae,0) and (-ae,0), we get
ae=4
Now we will put the value of e as given in the question. On doing so, we get
$\begin{align}
& a\times \dfrac{1}{3}=4 \\
& \Rightarrow a=12 \\
\end{align}$
Now the eccentricity of an ellipse is given by the formula: $e=\sqrt{1-\dfrac{{{\left( \text{length of semi-minor axis} \right)}^{2}}}{{{\left( \text{length of semi-major axis} \right)}^{2}}}}$ . Using this formula, we get
\[e=\sqrt{1-\dfrac{{{b}^{2}}}{{{12}^{2}}}}\]
Now using the value of e given in the question.
\[\dfrac{1}{3}=\sqrt{1-\dfrac{{{b}^{2}}}{{{12}^{2}}}}\]
\[\Rightarrow \dfrac{1}{9}=1-\dfrac{{{b}^{2}}}{{{12}^{2}}}\]
\[\Rightarrow b=\sqrt{\dfrac{8}{9}\times 144}=8\sqrt{2}\]
Therefore, the equation of the ellipse is:
$\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$
$\Rightarrow \dfrac{{{x}^{2}}}{144}+\dfrac{{{y}^{2}}}{128}=1$
Note: In mathematics, an ellipse is a plane curve surrounding two focal points, such that for all points on the curve, the sum of the two distances to the focal points is a constant.For these types of questions diagram is key point to approach the solution and students should know the definitions and parameters of ellipse i.e foci,vertices,major and minor axes and its equations ,eccentricity and latus rectum to solve these types of questions.
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