Phoebe

2020-10-26

Describe one similarity and one difference between the graphs of $\frac{{x}^{2}}{9}-\frac{{y}^{2}}{1}=1\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\frac{{\left(x-3\right)}^{2}}{9}-\frac{{\left(y+3\right)}^{2}}{1}=1$.

gotovub

Given:
(a).$\frac{{x}^{2}}{9}-\frac{{y}^{2}}{1}=1$
(b).$\frac{{\left(x-3\right)}^{2}}{9}-\frac{{\left(y+3\right)}^{2}}{1}=1$
To describe: One similarity and one difference between both the graphs.
Concept:
The sandard form of the equation of a hyperbola with center
(h,k)and transverse axis parallel to x - axis is
$\frac{{\left(x-h\right)}^{2}}{{a}^{2}}-\frac{{\left(y-k\right)}^{2}}{{b}^{2}}=1$, then
Lenght of transverse axis 2a.
Lenght of conjugate axis is 2b.
Distance between the foci is 2c, where ${c}^{2}={a}^{2}+{b}^{2}$.
Step 2
Explanation:
Here we have $\frac{{x}^{2}}{9}-\frac{{y}^{2}}{1}=1\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\frac{{\left(x-3\right)}^{2}}{9}-\frac{{\left(y+3\right)}^{2}}{1}=1$
i.e. $\frac{{x}^{2}}{{\left(3\right)}^{2}}-\frac{{y}^{2}}{{\left(1\right)}^{2}}=1\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\frac{{\left(x-3\right)}^{2}}{{\left(3\right)}^{2}}-\frac{{\left(y+3\right)}^{2}}{{\left(1\right)}^{2}}=1$
Difference: Both the hyperbola have different centre.
Center of $\frac{{x}^{2}}{{\left(3\right)}^{2}}-\frac{{y}^{2}}{{\left(1\right)}^{2}}=1$ is (0,0) while center of $\frac{{\left(x-3\right)}^{2}}{{\left(3\right)}^{2}}-\frac{{\left(y+3\right)}^{2}}{{\left(1\right)}^{2}}=1$ is (3,-3).
Similarity:Since both the hyperbola's have the same value of $\alpha =3\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}b=1$
$⇒$ Both the parabola's have same
lenght of transverse axes $=2a$
lenght of conjugate axes $=2b$
and distance between the foci $=2c.$

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