Question

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

• A. \(|t|<1\)
• B. \(|t|>1\)
• C. \(|t|=1 \)
• D. \(t∈(1, ∞] \)
• E. \(t∈(-∞,-1 ]\)

Answer 1

Answer: (E)

Given that;

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

So, let  us write the standard form as ;

\(Ax^2 + By^2 + Cx + Dy + E = 0\)

For the given equation:

\((1+t)x^2 + (t-1)y^2 + t^2 - 1 = 0\)

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

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. \((1+t≠0)\) and \((1-t≠0)\) for this case.)

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