If t be any non-zero number , then the locus of extremities of the family of hyperbolas . `t^(2)x^(2) - y^(2) = a^(2)t^(2)` is/are

Options

A. `x^(2) = a^(2) + ay`
B. `y^(2) = a^(2) + ax`
C. `x^(2) = a^(2) - ay`
D. `y^(2) = a^(2) - ax`

Answer

To find the locus of the extremities of the family of hyperbolas given by the equation \( t^2 x^2 - y^2 = a^2 t^2 \), we can follow these steps:

### Step 1: Rewrite the Equation
Start with the given equation:
\[ t^2 x^2 - y^2 = a^2 t^2 \]
Rearranging gives:
\[ t^2 x^2 = y^2 + a^2 t^2 \]

### Step 2: Divide by \( a^2 t^2 \)
To standardize the equation, divide both sides by \( a^2 t^2 \):
\[ \frac{t^2 x^2}{a^2 t^2} - \frac{y^2}{a^2 t^2} = 1 \]
This simplifies to:
\[ \frac{x^2}{a^2} - \frac{y^2}{a^2 t^2} = 1 \]

### Step 3: Identify the Parameters
From the standard form of the hyperbola, we can identify:
- The semi-major axis \( a \)
- The semi-minor axis \( b = a t \)

### Step 4: Find the Locus of Extremities
The extremities of the hyperbola occur at \( y = 0 \). Substitute \( y = 0 \) into the equation:
\[ \frac{x^2}{a^2} - 0 = 1 \implies x^2 = a^2 \implies x = \pm a \]
Thus, the extremities are at the points \( (a, 0) \) and \( (-a, 0) \).

### Step 5: Generalize for All \( t \)
Since \( t \) is a non-zero parameter, we can express the locus of these extremities as:
\[ x^2 = a^2 + a^2 t^2 \]
This can be rewritten as:
\[ x^2 = a^2(1 + t^2) \]
Let \( k = 1 + t^2 \), then:
\[ x^2 = a^2 k \]
This indicates that as \( t \) varies, the locus of the extremities forms a parabola.

### Final Equation
Thus, the locus of the extremities of the family of hyperbolas is given by:
\[ x^2 = a^2 + aky \]
This is the equation of a parabola.