Let the tangent to the parabola y²=12 x at the point (3, α) be perpendicular to the line 2 x+2 y=3. Then the square of distance of the point (6,-4) from the normal to the hyperbola α² x²-9 y²=9 α² at its point (α-1, α+2) is equal to _________.

Solution

Maths sir · 4 steps

Use the perpendicular condition to fix α = 6, then find the normal at (5, 8) on the hyperbola.

1. Find alpha. Since (3, α) lies on y² = 12x, we get α² = 36, so α = ± 6. Differentiating gives the tangent slope dy/dx = 6/y = 6/α. The line 2x + 2y = 3 has slope -1, so the tangent must have slope 1. That gives α = 6.
2. Set up hyperbola. With α = 6 the hyperbola is 36x² - 9y² = 324, that is x²/9 - y²/36 = 1. The point is (5, 8), and it lies on the curve because 25/9 - 64/36 = 1.
3. Normal at (5, 8). Differentiate 36x² - 9y² = 324 to get the tangent slope dy/dx = 4x/y = 20/8 = 5/2. The normal slope is therefore -2/5, so the normal is y - 8 = -2/5(x - 5), which simplifies to 2x + 5y = 50.

   d = (|2(6) + 5(-4) - 50|)/√(2² + 5²) = 58/√29 = 2√29

4. So. The square of the distance is d² = 4 × 29 = 116.

The common slip: 29. Using the tangent 5x - 2y = 9 instead of the normal gives a squared distance of 29, not 116.

The idea to keep. When a parameter comes with a sign ambiguity, let the geometric condition pick the sign. For a normal, find the tangent slope by implicit differentiation and take its negative reciprocal.

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