Question

How do you find the standard form given: - $9{{x}^{2}}+4{{y}^{2}}-36x+24y+36=0$.

Answer 1

Take the terms containing the variable x together and the terms containing the variable y together. Leave the constant term as it is. Now, use the method of completing the square and form expression of the form ${{\left( x-a \right)}^{2}}+{{\left( y-b \right)}^{2}}=k$ where ‘a’, ‘b’ and ‘k’ are some arbitrary constant that will be formed by balancing the equation.

Complete step by step answer:
Here, we have been provided with the expression: - $9{{x}^{2}}+4{{y}^{2}}-36x+24y+36=0$ and we are asked to write it in the standard form. To do this we need to use the method of completing the square.
Now, taking the terms containing the variable x together, and the terms containing the variable y together while leaving the constant term as it is, we get,

& \Rightarrow 9{{x}^{2}}-36x+4{{y}^{2}}+24y+36=0 \\\ & \Rightarrow 9\left( {{x}^{2}}-4x \right)+4\left( {{y}^{2}}+6y \right)+36=0 \\\ & \Rightarrow 9\left( {{x}^{2}}-2\times 2\times x \right)+4\left( {{y}^{2}}+2\times 3\times y \right)+36=0 \\\ \end{aligned}$$ Now, adding 4 inside the bracket of terms containing the variable x and balancing it by subtracting 36 in the overall equation, we have, $$\begin{aligned} & \Rightarrow 9\left( {{x}^{2}}-2\times 2\times x+4 \right)-36+\left( {{y}^{2}}+2\times 3\times y \right)+36=0 \\\ & \Rightarrow 9{{\left( x-2 \right)}^{2}}+4\left( {{y}^{2}}+2\times 3\times y \right)=0 \\\ \end{aligned}$$ Similarly, adding 9 inside the bracket of terms containing the variable y and balancing it by subtracting 36 in the overall equation, we have, $$\begin{aligned} & \Rightarrow 9{{\left( x-2 \right)}^{2}}+4\left( {{y}^{2}}+2\times 3\times y+9 \right)-36=0 \\\ & \Rightarrow 9{{\left( x-2 \right)}^{2}}+4{{\left( y+3 \right)}^{2}}=36 \\\ \end{aligned}$$ Dividing both the sides with 36, we get, $$\Rightarrow \dfrac{9{{\left( x-2 \right)}^{2}}}{36}+\dfrac{4{{\left( y+3 \right)}^{2}}}{36}=1$$ Simplifying we get, $$\Rightarrow \dfrac{{{\left( x-2 \right)}^{2}}}{4}+\dfrac{{{\left( y+3 \right)}^{2}}}{9}=1$$ $$\Rightarrow \dfrac{{{\left( x-2 \right)}^{2}}}{{{2}^{2}}}+\dfrac{{{\left( y+3 \right)}^{2}}}{{{3}^{2}}}=1$$ Hence, the above expression represents the standard form of the given expression: - $$9{{x}^{2}}+4{{y}^{2}}-36x+24y+36=0$$. **Note:** One may note that the standard form of the provided equation that we have obtained is actually an equation of an oblique ellipse. Whenever we have to find the centre of an oblique hyperbola, we use this completing the square method to convert the expression in standard form and then find our desired coordinates. Do not forget to balance the expression while using the completing square method, otherwise you will get the wrong answer.