The foci of the hyperbola are (11,-7) and (1, -7).
Given
The equation of a hyperbola is;
[tex]\rm \dfrac{(x-6)^2}{16}-\dfrac{(y+7)^2}{9}= 1[/tex]
The standard equation of a horizontal hyperbola with center (h, k) is;
[tex]\rm \dfrac{(x-h)^2}{a^2} -\dfrac{(y-k)^2}{b^2}=1[/tex]
The focus will be (h ± c, k) where c² = a²+b².
On comparing with the standard equation of hyperbola the value of a is 4 and b is 3.
The value of h is 6 and k is -7.
Then,
[tex]\rm c^2=a^2+b^2\\\\c^2 = 16+9\\\\c^2 =25\\\\c =5[/tex]
Therefore,
The foci of the hyperbola are;
[tex]\rm h+c= 6+5=11\\\\h-c=6-5=1\\\\k=-7[/tex]
Hence, the foci of the hyperbola are (11,-7) and (1, -7).
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Answer:
A: The foci of the hyperbola are (11,-7) and (1, -7).
Step-by-step explanation:
Correct as you can see on edge 2022!
