Question

Eccentricity of the ellipse ${{x}^{2}}+2{{y}^{2}}-2x+3y+2=0$ is:
1. $\dfrac{1}{\sqrt{2}}$
2. $\dfrac{1}{2}$
3. $\dfrac{1}{2\sqrt{2}}$
4. $\dfrac{1}{\sqrt{3}}$

Answer 1

Hint: For solving this question you should know about the eccentricity of the ellipse. As we know, an ellipse can be defined as the set of points in a plane in which the sum of distances from two fixed points is constant. And the eccentricity of the ellipse is always less than 1, which means $e<1$. And the general equation of ellipse is denoted as $\dfrac{\sqrt{{{a}^{2}}-{{b}^{2}}}}{a}$.

Complete step-by-step solution:
According to the question it is asked to us to find the eccentricity of the ellipse which is given as ${{x}^{2}}+2{{y}^{2}}-2x+3y+2=0$. As we know, the ellipse is defined as the set of points in which the sum of the distances from two fixed points is constant. And if we do this in simple terms, then we can say that the distance from the fixed point in a plane bears a constant ratio less than the distance from the fixed-point in a plane. So, it is clear that the eccentricity of the ellipse is less than 1, that is $e<1$.
The general equation of ellipse is denoted as $\dfrac{\sqrt{{{a}^{2}}-{{b}^{2}}}}{a}$.
So, according to the question,
${{x}^{2}}+2{{y}^{2}}-2x+3y+2=0$
Now if we solve this, then we will get,
$\begin{align}
  & \Rightarrow {{\left( x-1 \right)}^{2}}-1+2\left[ {{y}^{2}}+\left( \dfrac{3}{2} \right)y+\dfrac{9}{16}-\dfrac{9}{16} \right]+2=0 \\
 & \Rightarrow \dfrac{{{\left( x-1 \right)}^{2}}}{\left( \dfrac{1}{8} \right)}+\left(\dfrac{{{\left( y+\dfrac{3}{4} \right)}^{2}}}{\left( \dfrac{1}{16} \right)} \right)=1 \\
 & e=\sqrt{1-\left( \dfrac{{{b}^{2}}}{{{a}^{2}}} \right)} \\
 & =\sqrt{1-\left( \dfrac{\dfrac{1}{16}}{\dfrac{1}{8}} \right)} \\
 & =\sqrt{1-\left( \dfrac{1}{2} \right)} \\
 & =\dfrac{1}{\sqrt{2}} \\
\end{align}$
Hence the correct answer is option 1.

Note: while solving this type of questions you should be careful about the value of eccentricity, because for every next figure the eccentricity is found by a new formula. So, it will always be different and the value of $e$ is also different, it can be greater than or equal to 1 also.