how do i write an equation in standard form by completing the square for $x^2 -9y^2-4x-18y=14$

Question

I'm really having trouble with completeing squares i can solve for circles and ellipses but i can't seem to understand hyperbolas or parabolas, help would be deeply appreciated.

Answer 1

If you are having trouble with completing the square for your equation, let us rewrite your equation in the form: $$ Ax^2+Bx+Cy^2+Dy=E. $$ First, we complete the square to this term only: $Ax^2+Bx$. Take a look this following step: $$Ax^2+Bx=A\left(x^2+\dfrac{B}{A}x\right)=A\left(x+\dfrac{B}{2A}\right)^2-A\left(\dfrac{B}{2A}\right)^2.\tag1$$ Similarly with completing the square to the term: $Cy^2+Dy$. $$Cy^2+Dy=C\left(y^2+\dfrac{D}{C}y\right)=C\left(y+\dfrac{D}{2C}\right)^2-C\left(\dfrac{D}{2C}\right)^2.\tag2$$ To answer your question, we have $x^2-4x$ and $-9y^2-18y$. Using $(1)$, we obtain $$ \begin{align} x^2-4x&=1\left(x^2+\dfrac{(-4)}{(1)}x\right)\\ &=1\left(x+\dfrac{(-4)}{2(1)}\right)^2-1\left(\dfrac{(-4)}{2(1)}\right)^2\\ &=(x-2)^2-(-2)^2\\ &=\color{blue}{(x-2)^2-4}.\tag3 \end{align} $$ Using $(2)$, we obtain $$ \begin{align} -9y^2-18y&=-9\left(y^2+\dfrac{(-18)}{(-9)}y\right)\\ &=-9\left(y+\dfrac{(-18)}{2(-9)}\right)^2-(-9)\left(\dfrac{(-18)}{2(-9)}\right)^2\\ &=-9(y+1)^2+9(1)^2\\ &=\color{red}{-9(y+1)^2+9}.\tag4 \end{align} $$ Now, plug in $(3)$ and $(4)$ to your equation. Thus $$ \begin{align} x^2-9y^2-4x-18y&=14\\ \color{blue}{x^2-4x}\color{red}{-9y^2-18y}&=14\\ \color{blue}{(x-2)^2-4}+\color{red}{-9(y+1)^2+9}&=14\\ (x-2)^2-4-9(y+1)^2+9&=14\\ (x-2)^2-9(y+1)^2+5&=14\\ (x-2)^2-9(y+1)^2+5\color{green}{-5}&=14\color{green}{-5}\\ \Large\color{blue}{\boxed{(x-2)^2-9(y+1)^2=9}} \end{align} $$ To learn how to complete the square in an easy way, I encourage you (dorjan) to learn from this site.

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$$\large\color{blue}{\text{# }\mathbb{Q.E.D.}\text{ #}}$$

Answer 2

$$x^2-2\cdot x\cdot2+2^2-9(y^2+2\cdot y\cdot1+1^2)=14+4-9$$

$$\iff(x-2)^2-\{3(y-1)\}^2=3^2$$