Question

How do I use completing the square to rewrite the equation of an ellipse in standard form?

Answer 1

Hint: We have given a generalized equation of an ellipse and, we have to convert the generalized equation into the standard equation of an ellipse.
The generalized equation of an ellipse is given as $a{x^2} + b{y^2} + cx + dy + e = 0$ and standard equation of an ellipse is given as \[\dfrac{{{{\left( {x - h} \right)}^2}}}{{{a^2}}} + \dfrac{{{{\left( {y - k} \right)}^2}}}{{{b^2}}} = 1\] where $\left( {h,k} \right)$ is the center of the ellipse and $2a$ , $2b$ is the length of the major and minor axis respectively.
In the method of completing the square, we add and subtract the square of half of the coefficient of $x$ .

Complete step-by-step solution:
Step 1: We have given a generalized equation of an ellipse as $a{x^2} + b{y^2} + cx + dy + e = 0$. To convert it into standard form, first we rewrite the above equation as $a{x^2} + cx + b{y^2} + dy + e = 0$.
Now take the coefficient of $x$ common from the terms containing variable $x$ and take coefficient of $y$ common from the terms containing variable $y$ , we get
$ \Rightarrow a\left( {{x^2} + \dfrac{c}{a}x} \right) + b\left( {{y^2} + \dfrac{d}{b}y} \right) + e = 0$
Step 2: Now we add and subtract the square of half of the coefficient of $x$ and $y$ , we get
$ \Rightarrow a\left( {{x^2} + \dfrac{c}{a}x + {{\left( {\dfrac{c}{{2a}}} \right)}^2} - {{\left( {\dfrac{c}{{2a}}} \right)}^2}} \right) + b\left( {{y^2} + \dfrac{d}{b}y + {{\left( {\dfrac{d}{{2b}}} \right)}^2} - {{\left( {\dfrac{d}{{2b}}} \right)}^2}} \right) + e = 0$
Step 3: now converting each bracket into perfect square and simplifying , we get
\[
   \Rightarrow a\left( {{x^2} + 2 \times x \times \dfrac{c}{{2a}} + {{\left( {\dfrac{c}{{2a}}} \right)}^2}} \right) - a \times {\left( {\dfrac{c}{{2a}}} \right)^2} + b\left( {{y^2} + 2 \times y \times \dfrac{d}{b} + {{\left( {\dfrac{d}{{2b}}} \right)}^2}} \right) - b \times {\left( {\dfrac{d}{{2b}}} \right)^2} + e = 0 \\
   \Rightarrow a{\left( {x + \dfrac{c}{{2a}}} \right)^2} + b{\left( {y + \dfrac{d}{{2b}}} \right)^2} - a \times {\left( {\dfrac{c}{{2a}}} \right)^2} - b \times {\left( {\dfrac{d}{{2b}}} \right)^2} + e = 0 \\
 \]
Step 4: Now take all the constant terms on to the right-hand side, we get
\[ \Rightarrow a{\left( {x + \dfrac{c}{{2a}}} \right)^2} + b{\left( {y + \dfrac{d}{{2b}}} \right)^2} = \dfrac{{{c^2}}}{{4a}} + \dfrac{{{d^2}}}{{4b}} - e\]
Step 5: Now divide the entire equation by the constant term of the right side, we get
\[ \Rightarrow \dfrac{{a{{\left( {x + \dfrac{c}{{2a}}} \right)}^2}}}{{\dfrac{{{c^2}}}{{4a}} + \dfrac{{{d^2}}}{{4b}} - e}} + \dfrac{{b{{\left( {y + \dfrac{d}{{2b}}} \right)}^2}}}{{\dfrac{{{c^2}}}{{4a}} + \dfrac{{{d^2}}}{{4b}} - e}} = 1\]
Step 6: Now rewriting the above equation into standard form, we get
\[ \Rightarrow \dfrac{{{{\left( {x + \dfrac{c}{{2a}}} \right)}^2}}}{{\dfrac{1}{a}\left( {\dfrac{{{c^2}}}{{4a}} + \dfrac{{{d^2}}}{{4b}} - e} \right)}} + \dfrac{{{{\left( {y + \dfrac{d}{{2b}}} \right)}^2}}}{{\dfrac{1}{b}\left( {\dfrac{{{c^2}}}{{4a}} + \dfrac{{{d^2}}}{{4b}} - e} \right)}} = 1\]
Now the above equation is the standard equation of the ellipse whose center is at $\left( { - \dfrac{c}{{2a}}, - \dfrac{b}{{2d}}} \right)$

Note: Before adding and subtracting the square of half the coefficient of the variable, remember that the coefficient of the square term of the variable must be $1$ .
In the standard equation of an ellipse, the coefficient of the numerator is one.