Answer :
Answer:
[tex]\mathbf{y^2 -z^2 =1- n^2}[/tex]
Step-by-step explanation:
Give that:
the surface equation is [tex]x^2 +y^2 -z^2 =1[/tex]
from the family of traces in x = n given that [tex]x^2 +y^2 -z^2 =1[/tex] , the equation can be represented as :
[tex]n^2 +y^2 -z^2 =1[/tex]
[tex]\mathbf{y^2 -z^2 =1- n^2}[/tex]
This represents a family of hyperbola for all values of n expects that n = ± 1
So, if n = ± 1,
Then
[tex]y^2 - z^2 = 0[/tex]
(y-z) (y+z) = 0
y = ± z
So, for n = ± 1, it is a pair of line for y = z, y = -z