How do I graph the equation $16{{x}^{2}}-9{{y}^{2}}+32x+18y-137=0$ a TI$-83$ ?
Answer
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Hint: From the given expression, we can tell that this expression would result in a hyperbola. A hyperbola is a conic which can be defined as the difference of distances between a set of points, which are present in a plane to two fixed points is a positive constant. The expression that is given to us in the question is not the standard equation of a hyperbola. So we have to do some manipulations and make it look the equation of that of a standard hyperbola’s equation to graph it on a calculator.
Complete step by step answer:
We all know the standard equation of a hyperbola is $\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$ .
We have to convert $16{{x}^{2}}-9{{y}^{2}}+32x+18y-137=0$into the form $\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$.
Let us start off by grouping some terms.
Let us group $16{{x}^{2}},32x$ together and $-9{{y}^{2}},18y$ together.
Upon doing so, we get the following :
$\begin{align}
& \Rightarrow 16{{x}^{2}}-9{{y}^{2}}+32x+18y-137=0 \\
& \Rightarrow 16{{x}^{2}}+32x-9{{y}^{2}}+18y-137=0 \\
\end{align}$
Now let us take a $16$ common in the first two terms and a $-9$ common in the third and fourth terms.
Upon doing so, we get the following :
$\begin{align}
& \Rightarrow 16{{x}^{2}}-9{{y}^{2}}+32x+18y-137=0 \\
& \Rightarrow 16{{x}^{2}}+32x-9{{y}^{2}}+18y-137=0 \\
& \Rightarrow 16\left( {{x}^{2}}+2x \right)-9\left( {{y}^{2}}-2y \right)-137=0 \\
\end{align}$
From the standard equation, we can notice that our $x,y$ terms are squares. So even in this expression, we have our $x,y$ terms as some form of whole squares.
We see there is $\left( {{x}^{2}}+2x \right)$. And this can be converted into ${{\left( x+1 \right)}^{2}}$ if a $1$ is added to $\left( {{x}^{2}}+2x \right)$.
And in the same way, we have $\left( {{y}^{2}}-2y \right)$. And this can be converted into ${{\left( y-1 \right)}^{2}}$ if a $1$ is added to $\left( {{y}^{2}}-2y \right)$.
So let us do that. Let us add $1$ to each of them.
Upon doing so, we get the following :
$\begin{align}
& \Rightarrow 16{{x}^{2}}-9{{y}^{2}}+32x+18y-137=0 \\
& \Rightarrow 16{{x}^{2}}+32x-9{{y}^{2}}+18y-137=0 \\
& \Rightarrow 16\left( {{x}^{2}}+2x \right)-9\left( {{y}^{2}}-2y \right)-137=0 \\
& \Rightarrow 16\left( {{x}^{2}}+2x+1 \right)-9\left( {{y}^{2}}-2y+1 \right)-137-16+9=0 \\
\end{align}$
When we are adding a $1$ to $\left( {{x}^{2}}+2x \right)$, we are indirectly adding a $16$ since a $16$ is multiplied to the entire term of $\left( {{x}^{2}}+2x \right)$. And in the same way, when we are adding a $1$ to $\left( {{y}^{2}}-2y \right)$, we are indirectly subtracting a $9$ since a $-9$ is multiplied to the entire term of $\left( {{y}^{2}}-2y \right)$. And that is why we are separately subtracting a $16$ and adding a $9$ to balance out the equation.
$\begin{align}
& \Rightarrow 16{{x}^{2}}-9{{y}^{2}}+32x+18y-137=0 \\
& \Rightarrow 16{{x}^{2}}+32x-9{{y}^{2}}+18y-137=0 \\
& \Rightarrow 16\left( {{x}^{2}}+2x \right)-9\left( {{y}^{2}}-2y \right)-137=0 \\
& \Rightarrow 16\left( {{x}^{2}}+2x+1 \right)-9\left( {{y}^{2}}-2y+1 \right)-137-16+9=0 \\
& \Rightarrow 16{{\left( x+1 \right)}^{2}}-9{{\left( y-1 \right)}^{2}}=144 \\
& \Rightarrow \dfrac{{{\left( x+1 \right)}^{2}}}{9}-\dfrac{{{\left( y-1 \right)}^{2}}}{16}=1. \\
\end{align}$
So now it does like the standard equation of the hyperbola.
To graph it , we have to convert it into a function of $x$.
\[\begin{align}
& \Rightarrow \dfrac{{{\left( x+1 \right)}^{2}}}{9}-\dfrac{{{\left( y-1 \right)}^{2}}}{16}=1. \\
& \Rightarrow \dfrac{{{\left( x+1 \right)}^{2}}}{9}-1=\dfrac{{{\left( y-1 \right)}^{2}}}{16} \\
& \Rightarrow \dfrac{16{{\left( x+1 \right)}^{2}}}{9}-16={{\left( y-1 \right)}^{2}} \\
& \Rightarrow \left( y-1 \right)=\pm \dfrac{4}{3}\left[ \sqrt{{{\left( x+1 \right)}^{2}}-9} \right] \\
& \Rightarrow f\left( x \right)=y=\pm \dfrac{4}{3}\left[ \sqrt{{{\left( x+1 \right)}^{2}}-9} \right]+1 \\
\end{align}\]
Now on our TI$-83$calculator, we are going to graph the following :
\[\Rightarrow f\left( x \right)=\left\{ \begin{matrix}
& {{y}_{1}}&=\dfrac{4}{3}\left[ \sqrt{{{\left( x+1 \right)}^{2}}-9} \right]+1 \\
& {{y}_{2}}&=-\left( \dfrac{4}{3}\left[ \sqrt{{{\left( x+1 \right)}^{2}}-9} \right]+1 \right)=-{{y}_{1}} \\
\end{matrix} \right.\]
Let us go through steps to graph on the calculator.
Step $1$ : Click on the start button on the calculator.
Step $2$: Press the button STAT PLOT($Y=$) which is present on the top left corner.
Step $3$: Now, we have to punch in our expression against ${{Y}_{1}}$ . To get the radical or the square root , press on $2ND$ which is a blue color button also present on the top left corner.
Step $4$: Now, just type in our expression. After this, we also have to type in the negative part so as to get the complete hyperbola.
Step $5$: To get that, go to ${{Y}_{2}}$ which is right below ${{Y}_{1}}$. Against ${{Y}_{2}}$ , just press the negative symbol (-) and the press on the button which says VARS. This is present towards the right side of the calculator.
Step $6$: After pressing that, on the top of the calculator screen, we can see two options namely VARS,Y-VARS adjacent to each other. We have to click on Y-VARS and then select Function which is the foremost thing on the list of $4$.
Step $7$: After pressing that, we can see that there is ${{Y}_{1}}$ present in Functions. Click on ${{Y}_{1}}$ .Now we can ${{Y}_{2}}=-{{Y}_{1}}$ on the calculator screen.
Step $8$: After this, press on the WINDOWS button to set to get a full hyperbola of the size of our choice. This step is optional.
Step $9$: Press graph to see the hyperbola.
Note: It is important to remember the standard equations of all kinds of hyperbolas to complete the question easily. We should also remember the standard equations of the other conics as well since the question can always be a mixture of concepts. We should be able to manipulate when a certain hyperbolic equation is not in its standard form. And the manipulations are to be done carefully or else the entire answer may go wrong. We should be very careful while graphing and also giving input to the calculators.
Complete step by step answer:
We all know the standard equation of a hyperbola is $\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$ .
We have to convert $16{{x}^{2}}-9{{y}^{2}}+32x+18y-137=0$into the form $\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$.
Let us start off by grouping some terms.
Let us group $16{{x}^{2}},32x$ together and $-9{{y}^{2}},18y$ together.
Upon doing so, we get the following :
$\begin{align}
& \Rightarrow 16{{x}^{2}}-9{{y}^{2}}+32x+18y-137=0 \\
& \Rightarrow 16{{x}^{2}}+32x-9{{y}^{2}}+18y-137=0 \\
\end{align}$
Now let us take a $16$ common in the first two terms and a $-9$ common in the third and fourth terms.
Upon doing so, we get the following :
$\begin{align}
& \Rightarrow 16{{x}^{2}}-9{{y}^{2}}+32x+18y-137=0 \\
& \Rightarrow 16{{x}^{2}}+32x-9{{y}^{2}}+18y-137=0 \\
& \Rightarrow 16\left( {{x}^{2}}+2x \right)-9\left( {{y}^{2}}-2y \right)-137=0 \\
\end{align}$
From the standard equation, we can notice that our $x,y$ terms are squares. So even in this expression, we have our $x,y$ terms as some form of whole squares.
We see there is $\left( {{x}^{2}}+2x \right)$. And this can be converted into ${{\left( x+1 \right)}^{2}}$ if a $1$ is added to $\left( {{x}^{2}}+2x \right)$.
And in the same way, we have $\left( {{y}^{2}}-2y \right)$. And this can be converted into ${{\left( y-1 \right)}^{2}}$ if a $1$ is added to $\left( {{y}^{2}}-2y \right)$.
So let us do that. Let us add $1$ to each of them.
Upon doing so, we get the following :
$\begin{align}
& \Rightarrow 16{{x}^{2}}-9{{y}^{2}}+32x+18y-137=0 \\
& \Rightarrow 16{{x}^{2}}+32x-9{{y}^{2}}+18y-137=0 \\
& \Rightarrow 16\left( {{x}^{2}}+2x \right)-9\left( {{y}^{2}}-2y \right)-137=0 \\
& \Rightarrow 16\left( {{x}^{2}}+2x+1 \right)-9\left( {{y}^{2}}-2y+1 \right)-137-16+9=0 \\
\end{align}$
When we are adding a $1$ to $\left( {{x}^{2}}+2x \right)$, we are indirectly adding a $16$ since a $16$ is multiplied to the entire term of $\left( {{x}^{2}}+2x \right)$. And in the same way, when we are adding a $1$ to $\left( {{y}^{2}}-2y \right)$, we are indirectly subtracting a $9$ since a $-9$ is multiplied to the entire term of $\left( {{y}^{2}}-2y \right)$. And that is why we are separately subtracting a $16$ and adding a $9$ to balance out the equation.
$\begin{align}
& \Rightarrow 16{{x}^{2}}-9{{y}^{2}}+32x+18y-137=0 \\
& \Rightarrow 16{{x}^{2}}+32x-9{{y}^{2}}+18y-137=0 \\
& \Rightarrow 16\left( {{x}^{2}}+2x \right)-9\left( {{y}^{2}}-2y \right)-137=0 \\
& \Rightarrow 16\left( {{x}^{2}}+2x+1 \right)-9\left( {{y}^{2}}-2y+1 \right)-137-16+9=0 \\
& \Rightarrow 16{{\left( x+1 \right)}^{2}}-9{{\left( y-1 \right)}^{2}}=144 \\
& \Rightarrow \dfrac{{{\left( x+1 \right)}^{2}}}{9}-\dfrac{{{\left( y-1 \right)}^{2}}}{16}=1. \\
\end{align}$
So now it does like the standard equation of the hyperbola.
To graph it , we have to convert it into a function of $x$.
\[\begin{align}
& \Rightarrow \dfrac{{{\left( x+1 \right)}^{2}}}{9}-\dfrac{{{\left( y-1 \right)}^{2}}}{16}=1. \\
& \Rightarrow \dfrac{{{\left( x+1 \right)}^{2}}}{9}-1=\dfrac{{{\left( y-1 \right)}^{2}}}{16} \\
& \Rightarrow \dfrac{16{{\left( x+1 \right)}^{2}}}{9}-16={{\left( y-1 \right)}^{2}} \\
& \Rightarrow \left( y-1 \right)=\pm \dfrac{4}{3}\left[ \sqrt{{{\left( x+1 \right)}^{2}}-9} \right] \\
& \Rightarrow f\left( x \right)=y=\pm \dfrac{4}{3}\left[ \sqrt{{{\left( x+1 \right)}^{2}}-9} \right]+1 \\
\end{align}\]
Now on our TI$-83$calculator, we are going to graph the following :
\[\Rightarrow f\left( x \right)=\left\{ \begin{matrix}
& {{y}_{1}}&=\dfrac{4}{3}\left[ \sqrt{{{\left( x+1 \right)}^{2}}-9} \right]+1 \\
& {{y}_{2}}&=-\left( \dfrac{4}{3}\left[ \sqrt{{{\left( x+1 \right)}^{2}}-9} \right]+1 \right)=-{{y}_{1}} \\
\end{matrix} \right.\]
Let us go through steps to graph on the calculator.
Step $1$ : Click on the start button on the calculator.
Step $2$: Press the button STAT PLOT($Y=$) which is present on the top left corner.
Step $3$: Now, we have to punch in our expression against ${{Y}_{1}}$ . To get the radical or the square root , press on $2ND$ which is a blue color button also present on the top left corner.
Step $4$: Now, just type in our expression. After this, we also have to type in the negative part so as to get the complete hyperbola.
Step $5$: To get that, go to ${{Y}_{2}}$ which is right below ${{Y}_{1}}$. Against ${{Y}_{2}}$ , just press the negative symbol (-) and the press on the button which says VARS. This is present towards the right side of the calculator.
Step $6$: After pressing that, on the top of the calculator screen, we can see two options namely VARS,Y-VARS adjacent to each other. We have to click on Y-VARS and then select Function which is the foremost thing on the list of $4$.
Step $7$: After pressing that, we can see that there is ${{Y}_{1}}$ present in Functions. Click on ${{Y}_{1}}$ .Now we can ${{Y}_{2}}=-{{Y}_{1}}$ on the calculator screen.
Step $8$: After this, press on the WINDOWS button to set to get a full hyperbola of the size of our choice. This step is optional.
Step $9$: Press graph to see the hyperbola.
Note: It is important to remember the standard equations of all kinds of hyperbolas to complete the question easily. We should also remember the standard equations of the other conics as well since the question can always be a mixture of concepts. We should be able to manipulate when a certain hyperbolic equation is not in its standard form. And the manipulations are to be done carefully or else the entire answer may go wrong. We should be very careful while graphing and also giving input to the calculators.
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