To find the vertices and foci of the conic section: (x − 4)252 − (y + 3)262=1

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To find the vertices and foci of the conic section: (x − 4)252 − (y + 3)262=1

Velsenw · Accepted Answer

Concept used: Write the general equation of the hyperbola. (x−h)2a2−(y−k)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, (x−45)2−(y+36)2=1 (x−4)252−(y+3)262=1 Write the general equation of the hyperbola. (x−h)2a2−(y−k)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)(4−5,−3) =(9,−3)(−1,−3) The foci of the conic section are, (h±a,k)=(4±61,−3) =(4+61,−3)(4−61,−3) Therefore, the vertices and foci of the conic section are (9,−3)(−1,−3) and (4+61,−3)(4−61,−3) respectively.

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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