Question

How do you write the equation of the ellipse $9{{x}^{2}}+4{{y}^{2}}-72x+40y+208=0$ in standard form?

Answer 1

Hint: First group the terms containing the variable x together. Similarly group the terms containing the variable y together. Now, use completing the square method to convert the equation in the form ${{\left( \dfrac{x-\alpha }{a} \right)}^{2}}+{{\left( \dfrac{y-\beta }{b} \right)}^{2}}=1$, which is the standard form of the ellipse, to get the answer.

Complete step-by-step solution:
Here we have been provided with the expression of ellipse $9{{x}^{2}}+4{{y}^{2}}-72x+40y+208=0$ and we are asked to write its standard form. But first we need to know the standard equation of the ellipse.
Now, we know that the standard equation of an ellipse is given as ${{\left( \dfrac{x-\alpha }{a} \right)}^{2}}+{{\left( \dfrac{y-\beta }{b} \right)}^{2}}=1$. So we need to use completing the square method to get the desired expression. Since we have,
$\Rightarrow 9{{x}^{2}}+4{{y}^{2}}-72x+40y+208=0$
On grouping the terms containing the variable x together and similarly grouping the terms containing the variable y together we get,
$\Rightarrow \left( 9{{x}^{2}}-72x \right)+\left( 4{{y}^{2}}+40y \right)+208=0$
Now, taking 9 common from the first terms of x and 4 common from the terms of y, we can write the above expression as:
\[\begin{align}
  & \Rightarrow 9\left( {{x}^{2}}-8x \right)+4\left( {{y}^{2}}+10y \right)+208=0 \\
 & \Rightarrow 9\left( {{x}^{2}}-2\times x\times 4 \right)+4\left( {{y}^{2}}+2\times y\times 5 \right)+208=0 \\
\end{align}\]
Here we can clearly see that we need to add 16 inside the bracket of terms containing x and 25 inside the bracket of terms containing y to get the relation of the form $\left( {{x}^{2}}-2\alpha x+{{\alpha }^{2}} \right)+\left( {{y}^{2}}-2\beta y+{{\beta }^{2}} \right)$. To balance this change we need to subtract the terms added as a whole, so performing this arithmetic operation we get,
\[\begin{align}
  & \Rightarrow 9\left( {{x}^{2}}-2\times x\times 4+16 \right)+4\left( {{y}^{2}}+2\times y\times 5+25 \right)+208-\left( 9\times 16 \right)-\left( 4\times 25 \right)=0 \\
 & \Rightarrow 9\left( {{x}^{2}}-2\times x\times 4+16 \right)+4\left( {{y}^{2}}+2\times y\times 5+25 \right)+208-144-100=0 \\
 & \Rightarrow 9\left( {{x}^{2}}-2\times x\times 4+16 \right)+4\left( {{y}^{2}}+2\times y\times 5+25 \right)+208-244=0 \\
 & \Rightarrow 9\left( {{x}^{2}}-2\times x\times 4+16 \right)+4\left( {{y}^{2}}+2\times y\times 5+25 \right)-36=0 \\
\end{align}\]
Using the algebraic identity ${{a}^{2}}-2ab+{{b}^{2}}={{\left( a-b \right)}^{2}}$ and ${{a}^{2}}+2ab+{{b}^{2}}={{\left( a+b \right)}^{2}}$ we get,
\[\Rightarrow 9{{\left( x-4 \right)}^{2}}+4{{\left( y+5 \right)}^{2}}=36\]
Dividing both the sides with 36 we get,
\[\begin{align}
  & \Rightarrow \dfrac{9{{\left( x-4 \right)}^{2}}}{36}+\dfrac{4{{\left( y+5 \right)}^{2}}}{36}=1 \\
 & \Rightarrow \dfrac{{{\left( x-4 \right)}^{2}}}{4}+\dfrac{{{\left( y+5 \right)}^{2}}}{9}=1 \\
 & \Rightarrow \dfrac{{{\left( x-4 \right)}^{2}}}{{{2}^{2}}}+\dfrac{{{\left( y+5 \right)}^{2}}}{{{3}^{2}}}=1 \\
 & \Rightarrow {{\left( \dfrac{x-4}{2} \right)}^{2}}+{{\left( \dfrac{y+5}{3} \right)}^{2}}=1 \\
\end{align}\]
Hence, the above relation is our answer.

Note: You must not get confused in the standard equation of ellipse and do not consider it as ${{\left( \dfrac{x}{a} \right)}^{2}}+{{\left( \dfrac{y}{b} \right)}^{2}}=1$ because this is a particular case of ellipse in which the centre of ellipse lies on the origin. Now, you must be careful while using the square method. Never forget to subtract the terms which are added in extra otherwise the expression will get unbalanced and you will get the wrong answer.