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)

We are given the equation:

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

To determine when this represents a hyperbola, we analyze the coefficients of \(x^2\) and \(y^2\):

\[ A = 1 + t \quad \text{and} \quad B = t - 1 \]

A hyperbola occurs when \(A\) and \(B\) have opposite signs, i.e., \(A \cdot B < 0\). Thus:

\[ (1 + t)(t - 1) < 0 \]

Simplify the inequality:

\[ (1 + t)(t - 1) = t^2 - 1 < 0 \]

\[ t^2 < 1 \]

\[ |t| < 1 \]

Therefore, the equation represents a hyperbola when \(|t| < 1\).

The correct answer is:

(E) \(|t| < 1\)

Answer 2

Step 1: Analyze the given equation.

The given equation is:

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

We aim to determine for which values of \( t \) this equation represents a hyperbola.

Step 2: General condition for a conic section to be a hyperbola.

The general form of a second-degree equation in two variables is:

\[ Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0. \]

The discriminant of the conic section is given by:

\[ \Delta = B^2 - 4AC. \]

If \( \Delta > 0 \), the conic is a hyperbola.

In our case, there is no \( xy \)-term (\( B = 0 \)), so the discriminant simplifies to:

\[ \Delta = -4AC, \]

where \( A = 1 + t \) and \( C = t - 1 \).

Step 3: Compute the discriminant.

Substitute \( A = 1 + t \) and \( C = t - 1 \):

\[ \Delta = -4(1 + t)(t - 1). \]

Simplify:

\[ \Delta = -4((1)(t - 1) + t(t - 1)) = -4(t - 1 + t^2 - t) = -4(t^2 - 1). \]

Step 4: Determine when \( \Delta > 0 \).

For the conic to represent a hyperbola, we require \( \Delta > 0 \):

\[ -4(t^2 - 1) > 0. \]

Divide through by \(-4\) (reversing the inequality):

\[ t^2 - 1 < 0. \]

Factorize:

\[ (t - 1)(t + 1) < 0. \]

Step 5: Solve the inequality.

The product \((t - 1)(t + 1)\) is negative when \( t \) lies between \(-1\) and \( 1 \):

\[ -1 < t < 1. \]

Step 6: Interpret the result.

The equation represents a hyperbola if and only if \( |t| < 1 \).

Final Answer:

\( |t| < 1 \)