What are the foci of the hyperbola whose equation is = 1? (1,−7) and (11,−7) (2,−7) and (10,−7) (6,−12) and (6,−2) (6,−10) and (

Question

3 years ago

5

6,−4).

1 answer:

docker41 [41] · Answer 1 · 2022-10-28T09:33:40+03:00

The foci of the hyperbola are (11,-7) and (1, -7).

Given

The equation of a hyperbola is;

$\rm \dfrac{(x-6)^2}{16}-\dfrac{(y+7)^2}{9}= 1$

<h3>What is the standard equation of hyperbola?</h3>

The standard equation of a horizontal hyperbola with center (h, k) is;

$\rm \dfrac{(x-h)^2}{a^2} -\dfrac{(y-k)^2}{b^2}=1$

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,

$\rm c^2=a^2+b^2\\\\c^2 = 16+9\\\\c^2 =25\\\\c =5$

Therefore,

The foci of the hyperbola are;

$\rm h+c= 6+5=11\\\\h-c=6-5=1\\\\k=-7$

Hence, the foci of the hyperbola are (11,-7) and (1, -7).

To know more about Hyperbola click the link given below.