Question

Find the eccentricity of the ellipse whose equation is \[{{\left( x-3 \right)}^{2}}+{{\left( y-4 \right)}^{2}}=\dfrac{{{y}^{2}}}{9}\].
Choose the correct option:
A. \[\dfrac{\sqrt{3}}{2}\]
B. \[\dfrac{1}{3}\]
C. \[\dfrac{1}{3\sqrt{2}}\]
D. \[\dfrac{1}{\sqrt{3}}\]

Answer 1

Hint: The ellipse when written in the standard form \[\dfrac{{{\left( x-h \right)}^{2}}}{{{a}^{2}}}+\dfrac{{{\left( y-k \right)}^{2}}}{{{b}^{2}}}=1\], will have the eccentricity (e) given as\[e=\dfrac{\sqrt{{{b}^{2}}-{{a}^{2}}}}{b}\]if b is the major axis.

Complete step-by-step solution -
In the question, we have to find the eccentricity of the ellipse\[{{\left( x-3 \right)}^{2}}+{{\left( y-4 \right)}^{2}}=\dfrac{{{y}^{2}}}{9}\].
Now, we know that the standard form of the ellipse is written in the form of\[\dfrac{{{\left( x-h \right)}^{2}}}{{{a}^{2}}}+\dfrac{{{\left( y-k \right)}^{2}}}{{{b}^{2}}}=1\], whose major axis is b if \[b>a\]. So when b is the major axis, then the eccentricity of the ellipse will be given by the formula\[e=\dfrac{\sqrt{{{b}^{2}}-{{a}^{2}}}}{b}\].
Now, the ellipse that we have is \[{{\left( x-3 \right)}^{2}}+{{\left( y-4 \right)}^{2}}=\dfrac{{{y}^{2}}}{9}\], which will be first be written in the standard form, as shown below:
\[\begin{align}
  & \Rightarrow {{\left( x-3 \right)}^{2}}+{{\left( y-4 \right)}^{2}}=\dfrac{{{y}^{2}}}{9} \\
 & \Rightarrow {{\left( x-3 \right)}^{2}}+{{\left( y-4 \right)}^{2}}-\dfrac{{{y}^{2}}}{9}=0 \\
\end{align}\]
Next, we will multiply and divide the expression \[{{\left( x-3 \right)}^{2}}\]and \[{{\left( y-4 \right)}^{2}}\] with 9, to get:
\[\Rightarrow \dfrac{{{\left( x-3 \right)}^{2}}\cdot \ 9}{9}+\dfrac{{{\left( y-4 \right)}^{2}}\cdot \ 9}{9}-\dfrac{{{y}^{2}}}{9}=0\]
Then, we have to expand the expression \[{{\left( x-3 \right)}^{2}}\]and \[{{\left( y-4 \right)}^{2}}\]using the formula \[{{(a-b)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}\] and we get:
\[\begin{align}
  & \Rightarrow \dfrac{\left( {{x}^{2}}-6x+9 \right)\cdot \ 9+\left( {{y}^{2}}-8y+16 \right)\cdot \ 9-{{y}^{2}}}{9}=0 \\
 & \Rightarrow \dfrac{9{{x}^{2}}-54x+81+9{{y}^{2}}-72y+144-{{y}^{2}}}{9}=0 \\
 & \Rightarrow \dfrac{9{{x}^{2}}-54x+8{{y}^{2}}+225-72y}{9}=0 \\
\end{align}\]
Next, multiply both the sides with 9 to get:
\[\begin{align}
  & \Rightarrow 9{{x}^{2}}-54x+8{{y}^{2}}+225-72y=0 \\
 & \Rightarrow 9{{x}^{2}}-54x+8{{y}^{2}}-72y=-225 \\
 & \Rightarrow 9\left( {{x}^{2}}-6x \right)+8\left( {{y}^{2}}-9y \right)=-225 \\
\end{align}\]
Then after simplifying, we will divide each side of the equation by 9, to get:
\[\Rightarrow \left( {{x}^{2}}-6x \right)+\dfrac{8}{9}\left( {{y}^{2}}-9y \right)=-\dfrac{225}{9}\]
Now, divide each side of the equation by 8, to get:
\[\Rightarrow \dfrac{1}{8}\left( {{x}^{2}}-6x \right)+\dfrac{1}{9}\left( {{y}^{2}}-9y \right)=-\dfrac{25}{8}\]
Adding \[\dfrac{9}{8}\]both the sides to get:
\[\Rightarrow \dfrac{1}{8}\left( {{x}^{2}}-6x+9 \right)+\dfrac{1}{9}\left( {{y}^{2}}-9y \right)=-\dfrac{25}{8}+\dfrac{1}{8}\left( 9 \right)\]
Next, we need to add \[\dfrac{1}{9}\left( \dfrac{81}{4} \right)\] both the sides and simplify to get the expression of the form \[{{(a-b)}^{2}}\]. So, we get:
\[\begin{align}
  & \Rightarrow \dfrac{1}{8}{{\left( x-3 \right)}^{2}}+\dfrac{1}{9}{{\left( y-\dfrac{9}{2} \right)}^{2}}=-\dfrac{25}{8}+\dfrac{1}{8}\left( 9 \right)+\dfrac{1}{9}\left( \dfrac{81}{4} \right) \\
 & \Rightarrow \dfrac{1}{8}{{\left( x-3 \right)}^{2}}+\dfrac{1}{9}{{\left( y-\dfrac{9}{2} \right)}^{2}}=\dfrac{1}{4} \\
\end{align}\]
Next, we will divide both the sides of the equation by \[\dfrac{1}{4}\], to get:
\[\begin{align}
  & \Rightarrow \dfrac{{{\left( x-3 \right)}^{2}}}{2}+\dfrac{{{\left( y-\dfrac{9}{2} \right)}^{2}}}{\dfrac{9}{4}}=1 \\
 & \Rightarrow \dfrac{{{\left( x-3 \right)}^{2}}}{{{\left( \sqrt{2} \right)}^{2}}}+\dfrac{{{\left( y-\dfrac{9}{2} \right)}^{2}}}{{{\left( \dfrac{3}{2} \right)}^{2}}}=1 \\
\end{align}\]
Now, this is the standard form of the ellipse that has \[a=\sqrt{2},\;\,\,b=\dfrac{3}{2}\].
Here, \[b>a\]so the major axis is parallel to the y-axis. Thus the eccentricity (e) will be given as:
\[\begin{align}
  & \Rightarrow e=\dfrac{\sqrt{{{b}^{2}}-{{a}^{2}}}}{b} \\
 & \Rightarrow e=\dfrac{\sqrt{{{\left( \dfrac{3}{2} \right)}^{2}}-{{(\sqrt{2})}^{2}}}}{\left( \dfrac{3}{2} \right)} \\
 & \Rightarrow e=\dfrac{\sqrt{{{\left( \dfrac{3}{2} \right)}^{2}}-{{\left( \sqrt{2} \right)}^{2}}}\cdot \;2}{3} \\
 & \Rightarrow e=\dfrac{2\cdot \sqrt{\dfrac{9}{4}-2}}{3} \\
 & \Rightarrow e=\dfrac{2\cdot \dfrac{1}{2}}{3} \\
 & \Rightarrow e=\dfrac{1}{3} \\
\end{align}\]
Finally, the correct answer is option B) \[\dfrac{1}{3}\]



Note: It is important to simplify the equation so as to bring it in the standard form of the ellipse and care has to be taken while simplifying. The formula of the eccentricity depends on the major and the minor axis. Also, care has to be taken when finding the minor and the major axis of the ellipse, so as to decide the orientation of the ellipse.