SOLUTION
From the question, the center of the hyperbola is
a is the distance between the center to vertex, which is -1 or 1, and
c is the distance between the center to foci, which is -2 or 2.
b is given as
![\begin{gathered} b^2=c^2-a^2 \\ b^2=2^2-1^2 \\ b=\sqrt[]{3} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20b%5E2%3Dc%5E2-a%5E2%20%5C%5C%20b%5E2%3D2%5E2-1%5E2%20%5C%5C%20b%3D%5Csqrt%5B%5D%7B3%7D%20%5Cend%7Bgathered%7D)
But equation of a hyperbola is given as
Substituting the values of a, b, h and k, we have
![\begin{gathered} \frac{(x-0)^2}{1^2}-\frac{(y-0)^2}{\sqrt[]{3}^2}=1 \\ \frac{x^2}{1}-\frac{y^2}{3}=1 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20%5Cfrac%7B%28x-0%29%5E2%7D%7B1%5E2%7D-%5Cfrac%7B%28y-0%29%5E2%7D%7B%5Csqrt%5B%5D%7B3%7D%5E2%7D%3D1%20%5C%5C%20%5Cfrac%7Bx%5E2%7D%7B1%7D-%5Cfrac%7By%5E2%7D%7B3%7D%3D1%20%5Cend%7Bgathered%7D)
Hence the answer is
The given plane, 
, has normal vector 
. Any plane parallel to this one has the same normal vector.
Let 
be any point in the plane we want. The plane contains the point (1, 1, -1), so an arbitrary vector in this plane is
and this is perpendicular to 
.
So the equation of the plane is
or equivalently,
Answer:
y = 3/5 x + 2.
Step-by-step explanation:
Use the point-slope equation of a straight line:
y - y1 = m (x - x1) where m = the slope amd (x1, y1) is a point on the line.
Here:
m = (5- (-1)) / (5 - (-5))
= 6/10
= 3/5.
Substituting for m and (5, 5):
y - 5 = 3/5(x - 5)
y - 5 = 3/5x - 3
y = 3/5 x + 2.
