Question

Vertices of an ellipse are $\left( 0,\pm 10 \right)$ and its eccentricity is $e=\dfrac{4}{5}$ then the equation of ellipse is
(a) $90{{x}^{2}}-40{{y}^{2}}=3600$
(b) $80{{x}^{2}}+50{{y}^{2}}=4000$
(c) $36{{x}^{2}}+100{{y}^{2}}=3600$
(d) $100{{x}^{2}}+36{{y}^{2}}=3600$

Answer 1

We solve this problem by using the standard equation of the ellipse because the given vertices of the ellipse lie on the Y-axis. The rough figure of the ellipse is shown below.

The standard equation of ellipse having the co – ordinate axes as major and minor axis is $\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$ where ${{a}^{2}}={{b}^{2}}\left( 1-{{e}^{2}} \right)$
The above equation has vertices as $\left( 0,\pm b \right)$ because the major axis is the Y-axis.
By using the above equation we take the vertices of the ellipse to find the values of $a,b$

Complete step-by-step solution
We are given that the vertices of ellipse are $\left( 0,\pm 10 \right)$
We are also given that the eccentricity of ellipse as $e=\dfrac{4}{5}$
Let us assume that the equation of ellipse having the co – ordinate axes as major and minor axes of ellipse as
$\Rightarrow \dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$
We know that the vertices of above equation of ellipse are $\left( 0,\pm b \right)$
Now, by comparing the above vertices with given vertices we get
$\Rightarrow b=10$
Now, let us find the value of $'a'$
We know that for the standard equation of ellipse we have
${{a}^{2}}={{b}^{2}}\left( 1-{{e}^{2}} \right)$
By substituting the required values in above equation we get

& \Rightarrow {{a}^{2}}={{\left( 10 \right)}^{2}}\left( 1-{{\left( \dfrac{4}{5} \right)}^{2}} \right) \\\ & \Rightarrow {{a}^{2}}=100\left( 1-\dfrac{16}{25} \right) \\\ \end{aligned}$$ By subtracting the terms inside the brackets using the LCM we get $$\begin{aligned} & \Rightarrow {{a}^{2}}=100\left( \dfrac{25-16}{25} \right) \\\ & \Rightarrow {{a}^{2}}=4\times 9 \\\ & \Rightarrow {{a}^{2}}=36 \\\ \end{aligned}$$ Now, by substituting the values of $$'{{a}^{2}},b'$$ in the equation of ellipse we assumed we get $$\Rightarrow \dfrac{{{x}^{2}}}{36}+\dfrac{{{y}^{2}}}{100}=1$$ No, by adding the terms using the LCM we get $$\Rightarrow 100{{x}^{2}}+36{{y}^{2}}=3600$$ Therefore, the equation of ellipse having the vertices as $$\left( 0,\pm 10 \right)$$ and its eccentricity as $$e=\dfrac{4}{5}$$ is given as $$100{{x}^{2}}+36{{y}^{2}}=3600$$ **So, option (d) is the correct answer.** **Note:** Students may make mistakes in taking the equation of the ellipse. We are given that the vertices are $$\left( 0,\pm 10 \right)$$ this means that the major axis of the ellipse is the Y-axis. So, for the standard equation of ellipse $$\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$$ the vertices will be $$\left( 0,\pm b \right)$$ not $$\left( \pm a,0 \right)$$ Also the standard equation of ellipse $$\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$$ is applicable only when the major axis and minor axis are co – ordinate axes. These two points need to be taken care of.