Question

How do I convert the equation $4{{x}^{2}}-{{y}^{2}}-24x+4y+28=0$ to the standard form?

Answer 1

This question belongs to the topic of pre-calculus. In this question, first we will group the terms of x in one bracket and group the terms of y in another bracket. After that, we will add the terms to make the perfect square which are in the brackets and also we will subtract the terms outside the bracket to balance the equation. After that, we will solve the further equation and get the answer in standard form.

Complete step by step solution:
Let us solve this question.
In this question, we have asked to convert the given equation in the standard form. The given equation is $4{{x}^{2}}-{{y}^{2}}-24x+4y+28=0$. This equation can also be written as
$\Rightarrow \left( 4{{x}^{2}}-24x \right)-\left( {{y}^{2}}-4y \right)+28=0$
Now, we will add a third term in both the brackets to make a complete square of that bracket. We will add the term to make a perfect square in the form of ${{a}^{2}}-2\times a\times b+{{b}^{2}}$ in the first bracket and also we will add the term to make the perfect square in the form of ${{a}^{2}}-2\times a\times b+{{b}^{2}}$ in the second bracket. And, also we will subtract the terms to balance the equation.
So, we can write the above equation as
$\Rightarrow \left( 4{{x}^{2}}-24x+{{\left( \dfrac{24}{2\times 2} \right)}^{2}} \right)-\left( {{y}^{2}}-4y+{{\left( \dfrac{4}{2} \right)}^{2}} \right)+28-{{\left( \dfrac{24}{2\times 2} \right)}^{2}}-\left( -{{\left( \dfrac{4}{2} \right)}^{2}} \right)=0$
We can write the above equation as
$\Rightarrow \left( 4{{x}^{2}}-24x+36 \right)-\left( {{y}^{2}}-4y+4 \right)+28-36+4=0$
The above equation can also be written as
$\Rightarrow \left( {{\left( 2x \right)}^{2}}-2\times 2x\times 6+{{\left( 6 \right)}^{2}} \right)-\left( {{y}^{2}}-2\times y\times 2+{{\left( 2 \right)}^{2}} \right)+28-36+4=0$
The above equation can also be written as
$\Rightarrow \left( {{\left( 2x \right)}^{2}}-2\times 2x\times 12+{{\left( 6 \right)}^{2}} \right)-\left( {{y}^{2}}-2\times y\times 2+{{\left( 2 \right)}^{2}} \right)-4=0$
Using the formula ${{a}^{2}}-2\times a\times b+{{b}^{2}}={{\left( a-b \right)}^{2}}$, we can write the above equation as
$\Rightarrow {{\left( 2x-6 \right)}^{2}}-{{\left( y-2 \right)}^{2}}-4=0$
Now, taking common out 2 from the first bracket of the equation, we get
$\Rightarrow {{2}^{2}}\times {{\left( x-3 \right)}^{2}}-{{\left( y-2 \right)}^{2}}-4=0$
We can write the above equation as
$\Rightarrow 4{{\left( x-3 \right)}^{2}}-{{\left( y-2 \right)}^{2}}=4$
Now, dividing 4 to the both side of the equation, we get
$\Rightarrow {{\left( x-3 \right)}^{2}}-\dfrac{{{\left( y-2 \right)}^{2}}}{4}=1$
Hence, we get that the standard form of the equation $4{{x}^{2}}-{{y}^{2}}-24x+4y+28=0$ is ${{\left( x-3 \right)}^{2}}-\dfrac{{{\left( y-2 \right)}^{2}}}{4}=1$

Note:
As we can see that this question is from the topic of pre-calculus, so we should have a better knowledge in that topic for solving this type of question. We should remember the formula that is ${{a}^{2}}-2\times a\times b+{{b}^{2}}={{\left( a-b \right)}^{2}}$ to solve this type of question easily. We can see that this equation is in the form of hyperbola. The general equation of hyperbola is always in the form of $\dfrac{{{\left( x-h \right)}^{2}}}{{{a}^{2}}}-\dfrac{{{\left( y-k \right)}^{2}}}{{{b}^{2}}}=1$.