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Mathematics Question on Hyperbola

For a real number t ,the equation (1+t)x2+(t−1)y2+t2−1=0(1+t)x^2 + (t-1)y^2 + t^2 - 1 = 0 represents a hyperbola provided

A

∣t∣<1|t|<1

B

∣t∣>1|t|>1

C

∣t∣=1|t|=1

D

t∈(1,∞]t∈(1, ∞]

E

t∈(−∞,−1]t∈(-∞,-1 ]

Answer

t∈(−∞,−1]t∈(-∞,-1 ]

Explanation

Solution

Given that;

For a real number t ,the equation (1+t)x2+(t−1)y2+t2−1=0(1+t)x^2 + (t-1)y^2 + t^2 - 1 = 0, represents a hyperbola

So, let us write the standard form as ;

Ax2+By2+Cx+Dy+E=0Ax^2 + By^2 + Cx + Dy + E = 0

For the given equation:

(1+t)x2+(t−1)y2+t2−1=0(1+t)x^2 + (t-1)y^2 + t^2 - 1 = 0

We can see that A=1+t,B=t−1,C=0,D=0,A = 1+t, B = t-1, C = 0, D = 0, and E=t2−1.E = t^2 - 1.

For a conic section to be a hyperbola, it must satisfy the following conditions:

  1. (A and B have different signs).
  2. The coefficients of x^2 and y^2 are non-zero. (i.e. (1+t≠0)(1+t≠0) and (1−t≠0)(1-t≠0) for this case.)

Now after analyzing the options given we can state that; t∈(1,∞]t ∈ (1, ∞], satisfies the above two conditions and hence this is the answer.