Question

If the locus of the centroid of the triangle with vertices \( A(a,0) \), \( B(\cos t, a\sin t) \) and \( C(\sin t, -b\cos t) \) (\( t \) is a parameter) is given by \[ 9x^2 + 9y^2 - 6x = 49, \] then the area of the triangle formed by the line \[ \frac{x}{a} + \frac{y}{b} = 1 \] with the coordinate axes is: \

• A. \( \frac{49}{2} \)
• B. \( \frac{7}{2} \)
• C. \( 4 \)
• D. \( 47 \)
  \

Hint:

The area of a triangle formed by a straight line with the coordinate axes can be found using the formula \( \frac{1}{2} \times \text{x-intercept} \times \text{y-intercept} \).

Answer 1

The Correct Option is B

Step 1: Identifying the Locus Equation
The given locus equation of the centroid is: \[ 9x^2 + 9y^2 - 6x = 49. \] Rearrange this into the standard form of a circle: \[ 9(x^2 - \frac{2}{3}x) + 9y^2 = 49. \] Completing the square for \( x \): \[ 9 \left( x^2 - \frac{2}{3}x + \left(\frac{1}{3}\right)^2 \right) - 9 \left(\frac{1}{3}\right)^2 + 9y^2 = 49. \] \[ 9 \left( (x - \frac{1}{3})^2 \right) - \frac{9}{9} + 9y^2 = 49. \] \[ 9 (x - \frac{1}{3})^2 + 9y^2 = 50. \] Dividing by 9, \[ (x - \frac{1}{3})^2 + y^2 = \frac{50}{9}. \] Thus, the circle has center \( (\frac{1}{3}, 0) \) and radius \( \frac{\sqrt{50}}{3} \). Step 2: Finding the Area of the Triangle
The given equation of the line is: \[ \frac{x}{a} + \frac{y}{b} = 1. \] This line intersects the x-axis at \( (a, 0) \) and the y-axis at \( (0, b) \). The area of the triangle formed by this line with the coordinate axes is: \[ \frac{1}{2} \times a \times b. \] From the given equation, \[ \frac{x}{\frac{7}{3}} + \frac{y}{\frac{7}{3}} = 1. \] Comparing, \[ a = \frac{7}{3}, \quad b = \frac{7}{3}. \] Thus, the area of the triangle is: \[ \frac{1}{2} \times \frac{7}{3} \times \frac{7}{3} = \frac{7}{2}. \] Step 3: Conclusion
Thus, the final answer is: \[ \boxed{\frac{7}{2}}. \] \bigskip