Question

Write the eccentricity of the ellipse $9{x^2} + 5{y^2} - 18x - 2y - 16 = 0$.

Answer 1

We will convert the given equation in standard form by taking $9$ common from the terms of $x$ and $5$ from the terms of $y$. We will convert it into an equation similar to the standard equation given by: $\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1$, where $b > a$. The eccentricity of ellipse is: ${e^2} = 1 - \dfrac{{{a^2}}}{{{b^2}}}$. So, by putting the value of a and b, we can calculate the value of eccentricity.

Complete step-by-step answer:
We are given the equation of an ellipse as: $9{x^2} + 5{y^2} - 18x - 2y - 16 = 0$
We are required to calculate its eccentricity.
First of all, let us convert this equation similar to the standard equation of an ellipse given by: $\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1$, where $b > a$.
Taking $9$ common from the terms of $x$ and $5$ from the terms of $y$, we can write the equation as:
$\Rightarrow 9\left( {{x^2} - 2} \right) + 5\left( {{y^2} - \dfrac{2}{5}} \right) - 16 = 0$
Adding and subtracting ${1^2}$ from the first term and adding and subtracting ${\left( {\dfrac{1}{5}} \right)^2}$ in the second term of the equation, we get
$\Rightarrow 9\left( {{x^2} - 2 + {1^2}} \right) - 9\left( {{1^2}} \right) + 5\left( {{y^2} - \dfrac{2}{5} + \dfrac{1}{{{5^2}}}} \right) - 5\left( {\dfrac{1}{{{5^2}}}} \right) - 16 = 0$
On simplifying this equation, we get
$\Rightarrow 9\left( {{x^2} - 2 + {1^2}} \right) + 5\left( {{y^2} - \dfrac{2}{5} + \dfrac{1}{{{5^2}}}} \right) = 9\left( {{1^2}} \right) + 5\left( {\dfrac{1}{{{5^2}}}} \right) + 16$
Or, we can write this equation as:
$\Rightarrow 9\left( {{x^2} - 2 + {1^2}} \right) + 5\left( {{y^2} - \dfrac{2}{5} + \dfrac{1}{{{5^2}}}} \right) = 9 + \dfrac{1}{5} + 16$
$\Rightarrow 9\left( {{x^2} - 2 + {1^2}} \right) + 5\left( {{y^2} - \dfrac{2}{5} + \dfrac{1}{{{5^2}}}} \right) = \dfrac{{126}}{5}$
Now, using the formula: ${\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}$, we can write the above equation as:
$\Rightarrow 9{\left( {x - 1} \right)^2} + 5{\left( {y - \dfrac{1}{5}} \right)^2} = \dfrac{{126}}{5}$
Dividing both sides by $\dfrac{{126}}{5}$, we get
$\Rightarrow \dfrac{{9{{\left( {x - 1} \right)}^2}}}{{\dfrac{{126}}{5}}} + \dfrac{{5{{\left( {y - \dfrac{1}{5}} \right)}^2}}}{{\dfrac{{126}}{5}}} = 1$
Or, we can also write it as:
$\Rightarrow \dfrac{{{{\left( {x - 1} \right)}^2}}}{{\dfrac{{126}}{{45}}}} + \dfrac{{{{\left( {y - \dfrac{1}{5}} \right)}^2}}}{{\dfrac{{126}}{{25}}}} = 1$
This is similar to the standard equation $\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1$, where $b > a$.
Here, ${a^2} = \dfrac{{126}}{{45}}$ and ${b^2} = \dfrac{{126}}{{25}}$.
The eccentricity of an ellipse is given by the formula: ${e^2} = 1 - \dfrac{{{a^2}}}{{{b^2}}}$. On putting the values of ${a^2}$ and ${b^2}$, we get
$\Rightarrow {e^2} = 1 - \dfrac{{{a^2}}}{{{b^2}}} = 1 - \dfrac{{\dfrac{{126}}{{45}}}}{{\dfrac{{126}}{{25}}}}$
$\Rightarrow {e^2} = 1 - \dfrac{{25}}{{45}} = \dfrac{{45 - 25}}{{45}} = \dfrac{{20}}{{45}} = \dfrac{4}{9}$
Taking square roots both sides, we get
$\Rightarrow e = \dfrac{2}{3}$
Therefore, the eccentricity of the given ellipse is $\dfrac{2}{3}$.

Note: In this question, we may get confused in the steps while converting the given equation in terms of a standard equation of ellipse especially when we added and subtracted ${1^2}$ and $\dfrac{1}{{{5^2}}}$ to make it a perfect square. We can directly use the formula of eccentricity of the ellipse as: $e = \dfrac{1}{b}\sqrt {{b^2} - {a^2}}$ but it will be more complex since the taking the square root of $4$ and $9$ isn’t that tough.