Answer:
C)One of the vertices is (-10,2)
Step-by-step explanation:
Given the hyperbola: [tex]\dfrac{(x+5)^2}{4^2} -\dfrac{ (y-2)^2}{3^2} =1[/tex]
The standard equation for a hyperbola with a horizontal transverse axis is:
[tex]\dfrac{(x-h)^2}{a^2} -\dfrac{ (y-k)^2}{b^2} =1[/tex] where the center is at (h, k).
For our given hyperbola, the center (h,k)=(-5,2)
Now:
[tex]c^2=a^2+b^2\\c^2=4^2+3^2\\c^2=25\\c=5[/tex]
Since the center is at (-5,2), its foci (-c,0) and (c,0) are:
(-5-5,2) and (-5+5,2)= (-10,2) and (0,2)
Similarly, since the center is at (-5,2), its vertices, (-a,0) and (a,0) are:
(-5-4,2) and (-5+4,2)= (−9,2), (−1,2).
Therefore, the odd option is C.
