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

Question
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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