Given:

(a).\(\displaystyle\frac{{x}^{{2}}}{{9}}-\frac{{y}^{{2}}}{{1}}={1}\)

(b).\(\displaystyle\frac{{{\left({x}-{3}\right)}^{{2}}}}{{9}}-\frac{{{\left({y}+{3}\right)}^{{2}}}}{{1}}={1}\)

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

\(\displaystyle\frac{{{\left({x}-{h}\right)}^{{2}}}}{{a}^{{2}}}-\frac{{{\left({y}-{k}\right)}^{{2}}}}{{b}^{{2}}}={1}\), then

Lenght of transverse axis 2a.

Lenght of conjugate axis is 2b.

Distance between the foci is 2c, where \(\displaystyle{c}^{{2}}={a}^{{2}}+{b}^{{2}}\).

Step 2

Explanation:

Here we have \(\displaystyle\frac{{x}^{{2}}}{{9}}-\frac{{y}^{{2}}}{{1}}={1}{\quad\text{and}\quad}\frac{{{\left({x}-{3}\right)}^{{2}}}}{{9}}-\frac{{{\left({y}+{3}\right)}^{{2}}}}{{1}}={1}\)

i.e. \(\displaystyle\frac{{x}^{{2}}}{{{\left({3}\right)}^{{2}}}}-\frac{{y}^{{2}}}{{{\left({1}\right)}^{{2}}}}={1}{\quad\text{and}\quad}\frac{{{\left({x}-{3}\right)}^{{2}}}}{{{\left({3}\right)}^{{2}}}}-\frac{{{\left({y}+{3}\right)}^{{2}}}}{{{\left({1}\right)}^{{2}}}}={1}\)

Difference: Both the hyperbola have different centre.

Center of \(\displaystyle\frac{{x}^{{2}}}{{{\left({3}\right)}^{{2}}}}-\frac{{y}^{{2}}}{{{\left({1}\right)}^{{2}}}}={1}\) is (0,0) while center of \(\displaystyle\frac{{{\left({x}-{3}\right)}^{{2}}}}{{{\left({3}\right)}^{{2}}}}-\frac{{{\left({y}+{3}\right)}^{{2}}}}{{{\left({1}\right)}^{{2}}}}={1}\) is (3,-3).

Similarity:Since both the hyperbola's have the same value of \(\displaystyle\alpha={3}{\quad\text{and}\quad}{b}={1}\)

\(\displaystyle\Rightarrow\) Both the parabola's have same

lenght of transverse axes \(= 2a\)

lenght of conjugate axes \(= 2b\)

and distance between the foci \(= 2c.\)