Question

P is a variable point such that the distance of P from A(4,0) is twice the distance of P from B(-4,0). If the line \( 3y - 3x - 20 = 0 \) intersects the locus of P at the points C and D, then the distance between C and D is:

• A. \( 8 \)
• B. \( 8\sqrt{2} \)
• C. \( \frac{32}{3} \)
• D. \( 32 \)

Hint:

To solve problems involving loci and distances, first express the given conditions algebraically, then substitute the equation of the line into the locus equation and solve the resulting quadratic equation.

Answer 1

The Correct Option is C

Let the coordinates of point \( P \) be \( (x, y) \). The distance from \( P \) to \( A(4, 0) \) is: \[ (\text{Distance from } P { to } A = \sqrt{(x - 4)^2 + y^2}) \] The distance from \( P \) to \( B(-4, 0) \) is: \[ (\text{Distance from } P { to } B = \sqrt{(x + 4)^2 + y^2}) \] We are given that the distance from \( P \) to \( A \) is twice the distance from \( P \) to \( B \), so: \[ \sqrt{(x - 4)^2 + y^2} = 2 \times \sqrt{(x + 4)^2 + y^2} \] Squaring both sides: \[ (x - 4)^2 + y^2 = 4 \times \left[ (x + 4)^2 + y^2 \right] \] Expanding both sides: \[ (x^2 - 8x + 16) + y^2 = 4(x^2 + 8x + 16 + y^2) \] Simplifying the equation: \[ x^2 - 8x + 16 + y^2 = 4x^2 + 32x + 64 + 4y^2 \] \[ x^2 - 8x + 16 + y^2 - 4x^2 - 32x - 64 - 4y^2 = 0 \] \[ -3x^2 - 40x - 3y^2 - 48 = 0 \] Dividing by -3: \[ x^2 + \frac{40}{3}x + y^2 + 16 = 0 \] This is the equation of the locus of point \( P \). Step 2: The equation of the line \( 3y - 3x - 20 = 0 \) is: \[ y = x + \frac{20}{3} \] Substitute this into the equation of the locus of \( P \): \[ x^2 + \frac{40}{3}x + \left( x + \frac{20}{3} \right)^2 + 16 = 0 \] Expanding the square term: \[ x^2 + \frac{40}{3}x + \left( x^2 + \frac{40}{3}x + \frac{400}{9} \right) + 16 = 0 \] Combining like terms: \[ 2x^2 + \frac{80}{3}x + \frac{400}{9} + 16 = 0 \] Multiplying through by 9 to eliminate fractions: \[ 18x^2 + 240x + 400 + 144 = 0 \] \[ 18x^2 + 240x + 544 = 0 \] Divide by 8: \[ 9x^2 + 120x + 272 = 0 \] Now solve this quadratic equation using the quadratic formula: \[ x = \frac{-120 \pm \sqrt{120^2 - 4 \times 9 \times 272}}{2 \times 9} \] \[ x = \frac{-120 \pm \sqrt{14400 - 9792}}{18} \] \[ x = \frac{-120 \pm \sqrt{4608}}{18} \] \[ x = \frac{-120 \pm 67.89}{18} \] Thus, the two values of \( x \) are: \[ x_1 = \frac{-120 + 67.89}{18} = \frac{-52.11}{18} = -2.9 \] \[ x_2 = \frac{-120 - 67.89}{18} = \frac{-187.89}{18} = -10.4 \] The distance between \( C \) and \( D \) is: \[ |x_1 - x_2| = |-2.9 - (-10.4)| = 7.5 \] Thus, the distance between \( C \) and \( D \) is \( \frac{32}{3} \).