If A and B are the points of intersection of the circle x²+y²-8x=0 and the hyperbola x²/9-y²/4=1 and a point P moves on the line 2x-3y+4=0, then the centroid of Δ PAB lies on the line :

Solution

Maths sir · 4 steps

Find A and B first, then write P in terms of the centroid and put it back into P's line.

1. Eliminate y. The circle gives y² = 8x - x². The hyperbola is 4x² - 9y² = 36. Substituting gives 4x² - 72x + 9x² = 36, that is 13x² - 72x - 36 = 0.
2. Find A and B. The roots are x = (72 ± 84)/26, so x = 6 or x = -6/13. The second root gives y² = 8x - x² < 0, so reject it. At x = 6, y² = 48 - 36 = 12, so A = (6, 2√3) and B = (6, -2√3).
3. Centroid from P. Let P = (h, k), with 2h - 3k + 4 = 0. The centroid is ((12 + h)/3, k/3) = (X, Y). So h = 3X - 12 and k = 3Y.

   2(3X - 12) - 3(3Y) + 4 = 0 ⇒ 6X - 9Y = 20

4. So. The centroid of Δ PAB always lies on 6x - 9y = 20, which is option (d).

The idea to keep. For a locus, call the moving point (X, Y). Express the original variable point in terms of X and Y, then substitute into the condition it satisfies. Always check that each intersection root is real.

Answer from the NTA final answer key. Solution by PrepWiser.