I If a point `(alpha, beta)` lies on the circle `x^2 +y^2=1` then the locus of the point `(3alpha.+2, beta),` is

Options

A. a straight line
B. an ellipse
C. a parabola
D. none of these

Explanation

To find the locus of the point \((3\alpha - 2, \beta)\) given that the point \((\alpha, \beta)\) lies on the circle defined by the equation \(x^2 + y^2 = 1\), we can follow these steps:

### Step 1: Define the new coordinates
Let:
- \(H = 3\alpha - 2\)
- \(K = \beta\)

### Step 2: Express \(\alpha\) and \(\beta\) in terms of \(H\) and \(K\)
From the definition of \(H\):
\[ \alpha = \frac{H + 2}{3} \]
And since \(K = \beta\), we have:
\[ \beta = K \]

### Step 3: Substitute \(\alpha\) and \(\beta\) into the circle equation
Since \((\alpha, \beta)\) lies on the circle \(x^2 + y^2 = 1\), we substitute \(\alpha\) and \(\beta\) into this equation:
\[ \alpha^2 + \beta^2 = 1 \]
Substituting the expressions for \(\alpha\) and \(\beta\):
\[ \left(\frac{H + 2}{3}\right)^2 + K^2 = 1 \]

### Step 4: Simplify the equation
Expanding the left-hand side:
\[ \frac{(H + 2)^2}{9} + K^2 = 1 \]
Multiply through by 9 to eliminate the fraction:
\[ (H + 2)^2 + 9K^2 = 9 \]

### Step 5: Rearrange the equation
Rearranging gives:
\[ (H + 2)^2 + 9K^2 = 9 \]

### Step 6: Identify the locus
This equation represents an ellipse centered at \((-2, 0)\) with semi-major axis 3 (along the \(K\) direction) and semi-minor axis 1 (along the \(H\) direction).

### Final Equation
Thus, the locus of the point \((3\alpha - 2, \beta)\) is given by:
\[ \frac{(H + 2)^2}{9} + \frac{K^2}{1} = 1 \]