Question

How do you graph $\left( {\dfrac{{{x^2}}}{4}} \right) - \left( {\dfrac{{{y^2}}}{9}} \right) = 1$?

Answer 1

We use the definition of hyperbola and write its general equation. Compare the given equation to the general equation of hyperbola and write values of center and vertices.

• General equation of hyperbola having center $(h,k)$and vertices $(h, \pm a)$is given by the equation $\dfrac{{{{\left( {x - h} \right)}^2}}}{{{a^2}}} - \dfrac{{{{\left( {y - k} \right)}^2}}}{{{b^2}}} = 1$.

Complete step-by-step answer:
We are given the equation $\left( {\dfrac{{{x^2}}}{4}} \right) - \left( {\dfrac{{{y^2}}}{9}} \right) = 1$
We can write this equation in simpler form i.e. $\dfrac{{{{\left( {x - 0} \right)}^2}}}{{{2^2}}} - \dfrac{{{{\left( {y - 0} \right)}^2}}}{{{3^2}}} = 1$
On comparing this equation with general equation of hyperbola $\dfrac{{{{\left( {x - h} \right)}^2}}}{{{a^2}}} - \dfrac{{{{\left( {y - k} \right)}^2}}}{{{b^2}}} = 1$, we get
$h = 0;k = 0$
So, the center $(h,k)$ of the parabola becomes $(0,0)$ … (1)
Now from comparison we can also write $a = 2;b = 3$
So, the vertices $(h, \pm a)$ of the hyperbola become $(0, \pm 2)$
Asymptotes are the straight lines that continuously approach the curve but never meet the curve
Now we can construct the hyperbola $\left( {\dfrac{{{x^2}}}{4}} \right) - \left( {\dfrac{{{y^2}}}{9}} \right) = 1$ using its center $(0,0)$ and its vertices $(0, \pm 2)$
We know the hyperbola will open in left and right side as the positive value of ‘x’
We first mark the center of hyperbola on the graph and then take 2 units in upward, downward, left and right direction

Now connect these corners as a rectangle and draw lines through corners as asymptotes

Draw hyperbola using the asymptotes opened on right and left side

$\therefore$Graph of hyperbola $\left( {\dfrac{{{x^2}}}{4}} \right) - \left( {\dfrac{{{y^2}}}{9}} \right) = 1$ is

Note:
Many students make mistake of assuming the equation given in the question as equation of an ellipse as they think the general equation of ellipse is $\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1$ and they think negative sign just indicates graph is elliptical in which direction, but this is wrong concept. Keep in mind we never have a negative sign in an ellipse, a negative sign in such an equation tells us that the equation is of hyperbola.