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# To find the vertices and foci of the conic section: frac{(x - 4)^{2}}{5^{2}} - frac{(y + 3)^{2}}{6^{2}}=1 # To find the vertices and foci of the conic section: frac{(x - 4)^{2}}{5^{2}} - frac{(y + 3)^{2}}{6^{2}}=1

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Conic sections asked 2021-02-20
To find the vertices and foci of the conic section: $$\displaystyle{\frac{{{\left({x}\ -\ {4}\right)}^{{{2}}}}}{{{5}^{{{2}}}}}}\ -\ {\frac{{{\left({y}\ +\ {3}\right)}^{{{2}}}}}{{{6}^{{{2}}}}}}={1}$$

## Answers (1) 2021-02-21
Concept used: Write the general equation of the hyperbola. $$\displaystyle{\frac{{{\left({x}-{h}\right)}^{{{2}}}}}{{{a}^{{{2}}}}}}-{\frac{{{\left({y}-{k}\right)}^{{{2}}}}}{{{b}^{{{2}}}}}}={1}$$ The centre of the hyperbola is (h,k), the vertices are $$\displaystyle{\left({h}\pm{a},{k}\right)}$$, and foci are $$\displaystyle{\left({h}\pm{c},{k}\right)}$$ Write the expression for c. $$\displaystyle{c}=\sqrt{{{a}^{{{2}}}+{b}^{{{2}}}}}$$ The given equation is, $$\displaystyle{\left({\frac{{{x}-{4}}}{{{5}}}}\right)}^{{{2}}}-{\left({\frac{{{y}+{3}}}{{{6}}}}\right)}^{{{2}}}={1}$$
$$\displaystyle{\frac{{{\left({x}-{4}\right)}^{{{2}}}}}{{{5}^{{{2}}}}}}-{\frac{{{\left({y}+{3}\right)}^{{{2}}}}}{{{6}^{{{2}}}}}}={1}$$ Write the general equation of the hyperbola. $$\displaystyle{\frac{{{\left({x}-{h}\right)}^{{{2}}}}}{{{a}^{{{2}}}}}}-{\frac{{{\left({y}-{k}\right)}^{{{2}}}}}{{{b}}}}={1}$$ Compare the given equation with the general equation of the hyperbola to get, $$\displaystyle{h}={4}$$
$$\displaystyle{k}=-{3}$$
$$\displaystyle{a}={5}$$
$$\displaystyle{b}={6}$$ Write the expression for c. $$\displaystyle{c}=\sqrt{{{a}^{{{2}}}+{b}^{{{2}}}}}$$ Substitute 6 for a and 5 for b. $$\displaystyle{c}=\sqrt{{{5}^{{{2}}}+{6}^{{{2}}}}}$$
$$\displaystyle=\sqrt{{{25}+{36}}}$$
$$\displaystyle=\sqrt{{{61}}}$$ The vertices of the conic section are, $$\displaystyle{\left({h}\pm{a},{k}\right)}={\left({4}\pm{5},-{3}\right)}$$
$$\displaystyle={\left({4}+{5},-{3}\right)}{\left({4}-{5},-{3}\right)}$$
$$\displaystyle={\left({9},-{3}\right)}{\left(-{1},-{3}\right)}$$ The foci of the conic section are, $$\displaystyle{\left({h}\pm{a},{k}\right)}={\left({4}\pm\sqrt{{{61}}},-{3}\right)}$$
$$\displaystyle={\left({4}+\sqrt{{{61}}},-{3}\right)}{\left({4}-\sqrt{{{61}}},-{3}\right)}$$ Therefore, the vertices and foci of the conic section are $$\displaystyle{\left({9},-{3}\right)}{\left(-{1},-{3}\right)}$$ and PSK(4+\sqrt{61},-3)(4-\sqrt{61},-3) respectively.

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