Let the function f(x)=2 x²-log ₑ x, x>0, be decreasing in (0, a) and increasing in (a, 4). A tangent to the parabola y²=4 a x at a point P on it passes through the point (8 a, 8 a-1) but does not pass through the point (-1/a, 0). If the equation of the normal at P is : x/α+y/β=1, then α+β is equal to ________________.

Solution

Maths sir · 4 steps

First get a from the minimum of f; then two tangents pass through the point, and you keep only one.

1. Find a. f'(x) = 4x - 1/x, which is zero at x = 1/2 for x > 0. It is negative before this point and positive after it, so a = 1/2. The parabola is y² = 2x, the point is (4, 3) and the excluded point is (-2, 0).
2. Tangents through (4, 3). A tangent to y² = 4ax with slope m is y = mx + a/m, which here is y = mx + 1/2m. Putting in (4, 3) gives 3 = 4m + 1/2m, so 8m² - 6m + 1 = 0 and m = 1/2 or 1/4.

   Slope m	Tangent	Through (-2, 0)?	Point of contact (a/m², 2a/m)
   1/2	y = x/2 + 1	-1 + 1 = 0, yes, so rejected	(2, 2)
   1/4	y = x/4 + 2	-1/2 + 2 ≠ 0, no, so kept	(8, 4)

3. Normal at P. At P(8, 4) the tangent slope is 1/4, so the normal slope is -4. The normal is y - 4 = -4(x - 8), that is 4x + y = 36.

   x/9 + y/36 = 1

4. So. α = 9 and β = 36, so α + β = 45.

The common slip: 9. Using the rejected slope m = 1/2 gives P(2, 2) and the normal x/3 + y/6 = 1, so the total is 9, not 45.

The idea to keep. From an outside point there are always two tangents to a parabola. Use y = mx + a/m to get both slopes, then use the extra condition in the question to decide which one is meant.

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