Let P(x₀, y₀) be the point on the hyperbola 3 x²-4 y²=36, which is nearest to the line 3 x+2 y=1. Then √2(y₀-x₀) is equal to :

Solution

Maths sir · 4 steps

The nearest point is where the tangent runs parallel to the line, so match the slopes first.

1. Why parallel tangent. The line does not meet the hyperbola. Substituting y = (1-3x)/2 gives 6x² - 6x + 37 = 0, whose discriminant is negative. So the nearest point is where the tangent has the same slope as the line, -3/2.
2. Match the slopes. The tangent at (x₀, y₀) is 3x x₀ - 4y y₀ = 36, with slope 3x₀/4y₀. Setting this equal to -3/2 gives x₀ = -2y₀.

   3(4y₀²) - 4y₀² = 36 ⇒ y₀² = 9/2 ⇒ y₀ = ±3/√2

3. Pick the nearer one. On the line side, 3x₀ + 2y₀ - 1 = -4y₀ - 1. For y₀ = -3/√2 this is 6√2 - 1. For y₀ = +3/√2 it is -(6√2 + 1). The first is smaller in size, so y₀ = -3/√2 and x₀ = 3√2.
4. So. y₀ - x₀ = -3/√2 - 6/√2 = -9/√2. Multiplying by √2 gives √2(y₀ - x₀) = -9, which is option (b).

The idea to keep. For the point of a conic nearest to a line that misses it, find the tangents parallel to the line. You get two points of contact, so always check which one is closer.

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