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

Lennie Carroll 2021-02-20 Answered
To find the vertices and foci of the conic section: (x  4)252  (y + 3)262=1
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Expert Answer

Velsenw
Answered 2021-02-21 Author has 91 answers

Concept used: Write the general equation of the hyperbola. (xh)2a2(yk)2b2=1 The centre of the hyperbola is (h,k), the vertices are (h±a,k), and foci are (h±c,k) Write the expression for c. c=a2+b2 The given equation is, (x45)2(y+36)2=1
(x4)252(y+3)262=1 Write the general equation of the hyperbola. (xh)2a2(yk)2b=1 Compare the given equation with the general equation of the hyperbola to get, h=4
k=3
a=5
b=6 Write the expression for c. c=a2+b2 Substitute 6 for a and 5 for b. c=52+62
=25+36
=61 The vertices of the conic section are, (h±a,k)=(4±5,3)
=(4+5,3)(45,3)
=(9,3)(1,3) The foci of the conic section are, (h±a,k)=(4±61,3)
=(4+61,3)(461,3) Therefore, the vertices and foci of the conic section are (9,3)(1,3) and (4+61,3)(461,3) respectively.

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