Let the normals at all the points on a given curve pass through a fixed point (a, b). If the curve passes through (3, -3) and (4, -2√2), and given that a - 2√2 b = 3,
then (a² + b² + ab) is equal to __________.

Solution

Maths sir · 5 steps

If every normal passes through one point, the curve is a circle with that point as its centre.

1. Identify the curve. The normal at (x, y) passes through (a, b), so the tangent there is perpendicular to the line joining (x, y) to (a, b). That gives (x - a) + (y - b)dy/dx = 0. Integrating, (x-a)² + (y-b)² is constant, so the curve is a circle centred at (a, b).
2. Equal distances. Both given points are on the circle, so they are the same distance from the centre. Equate the squared distances.

   (3-a)² + (-3-b)² = (4-a)² + (-2√2-b)²

3. Simplify. The a² and b² terms cancel: 18 - 6a + 6b = 24 - 8a + 4√2 b. This gives a + (3 - 2√2) b = 3.
4. Use the given condition. Subtract a - 2√2 b = 3 from this equation: (3 - 2√2)b + 2√2 b = 0, so 3b = 0. Then b = 0 and a = 3. Check: both points are at distance 3 from (3, 0).
5. So. a² + b² + ab = 9 + 0 + 0 = 9.

The idea to keep. Normals all through one fixed point means a circle centred there. Then any two points on the curve are equidistant from that centre, and the a², b² terms cancel to leave a linear equation.

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