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 Hyperbola Equation
We start with the equation of the hyperbola:
\[ t^2 x^2 - y^2 = a^2 t^2 \]
We can rearrange this equation to the standard form of a hyperbola:
\[ \frac{x^2}{a^2} - \frac{y^2}{a^2 t^2} = 1 \]

### Step 2: Identify the Parameters
From the standard form, we can identify:
- \( a^2 \) is the square of the semi-major axis.
- \( b^2 = a^2 t^2 \) is the square of the semi-minor axis.

### Step 3: Find the Lattice Rectum
The length of the lattice rectum of a hyperbola is given by \( \frac{2b^2}{a} \). For our hyperbola:
\[ b^2 = a^2 t^2 \implies \text{Length of lattice rectum} = \frac{2(a^2 t^2)}{a} = 2at^2 \]

### Step 4: Determine the Extremities
The extremities of the hyperbola occur at points where \( x = ae \) and \( y = \pm b^2/a \). We can find the coordinates of the extremities:
- For \( x = ae \), we have \( y = \frac{b^2}{a} = \frac{a^2 t^2}{a} = at^2 \) and \( y = -at^2 \).

Thus, the extremities are:
1. \( (ae, at^2) \)
2. \( (ae, -at^2) \)

### Step 5: Substitute for \( t \)
To find the locus, we need to eliminate \( t \). We know:
\[ e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + t^2} \]
Thus, \( ae = a\sqrt{1 + t^2} \) implies:
\[ x = a\sqrt{1 + t^2} \]
Squaring both sides gives:
\[ x^2 = a^2(1 + t^2) \implies t^2 = \frac{x^2}{a^2} - 1 \]

### Step 6: Substitute \( t^2 \) into \( y \)
Now substituting \( t^2 \) into the expression for \( y \):
\[ y = at^2 = a\left(\frac{x^2}{a^2} - 1\right) = \frac{x^2}{a} - a \]

### Step 7: Final Locus Equation
Thus, the locus of the extremities can be expressed as:
\[ y = \frac{x^2}{a} - a \]

### Step 8: Consider Negative Extremities
For the negative extremities, we have:
\[ y = -at^2 = -a\left(\frac{x^2}{a^2} - 1\right) = -\frac{x^2}{a} + a \]

### Final Result
The equations for the locus of the extremities of the hyperbola are:
1. \( y = \frac{x^2}{a} - a \)
2. \( y = -\frac{x^2}{a} + a \)