Question

How do you graph $\dfrac{{{x^2}}}{{64}} - {y^2} = 1$ and identify the foci and asymptotes?

Answer 1

We have given an equation of a line as $\dfrac{{{x^2}}}{{64}} - {y^2} = 1$ , which is an equation of hyperbola. A standard hyperbolic equation is always represented as $\dfrac{{{{(x - h)}^2}}}{{{a^2}}} - \dfrac{{{{(y - k)}^2}}}{{{b^2}}} = 1$ where the foci of the hyperbola is given as $\left( {h \pm \sqrt {{a^2} + {b^2}} ,k} \right)$ and the equation of asymptotes are given as
$y = \pm \dfrac{b}{a}(x - h) + k$.

Complete step by step solution:
We have equation of hyperbola as,
$\dfrac{{{x^2}}}{{64}} - {y^2} = 1$
Now we compare this given hyperbolic equation with the standard hyperbolic equation i.e., $\dfrac{{{{(x - h)}^2}}}{{{a^2}}} - \dfrac{{{{(y - k)}^2}}}{{{b^2}}} = 1$
Hence , we observe that $a = 8,b = 1,$ and $h = k = 0$ .
Therefore, we can say that the values of foci are at :
$\left( {0 - \sqrt {{8^2} + {1^2}} ,0} \right)$ and $\left( {0 + \sqrt {{8^2} + {1^2}} ,0} \right)$ .
After simplifying we will get ,
$\left( { - \sqrt {65} ,0} \right)$ and $\left( {\sqrt {65} ,0} \right)$ .
Now, we can also say that the equation of asymptotes are:
$y = - \dfrac{1}{8}(x - 0) + 0$ and $y = \dfrac{1}{8}(x - 0) + 0$ .
After simplifying we will get ,
$y = - \dfrac{1}{8}x$ and $y = \dfrac{1}{8}x$ .
The graph is $\dfrac{{{x^2}}}{{64}} - {y^2} = 1$ $[ - 10,10, - 5,5]$ .

Additional Information:
Since the ‘y’ term of the given equation is negative and the ‘x’ term is the positive one , we can say that this must be an hourglass shaped graph and not the baseball shaped graph. The standard form of a hyperbola is as follows:
$\dfrac{{{{(x - h)}^2}}}{{{a^2}}} - \dfrac{{{{(y - k)}^2}}}{{{b^2}}} = 1$
Where :
‘h’ is representing the distance the graph is shifted from the y-axis ,
‘k’ is representing the distance the graph is shifted from the x-axis,
‘a’ is representing the distance from the vertex of the graph to the centre of the graph,
‘b’ is used to indicate the vertical stretch.

Note:
There exists a correlation between the variables as ${a^2} + {b^2} = {c^2}$ where ‘c’ is representing the distance from the foci of the graph to the centre of the graph. Therefore, ‘c’ will always be greater than ‘a’.