Let m₁ and m₂ be the slopes of the tangents drawn from the point P(4,1) to the hyperbola H: y²/25-x²/16=1. If Q is the point from which the tangents drawn to H have slopes |m₁| and |m₂| and they make positive intercepts α and β on the x-axis, then (P Q)²/(α β) is equal to __________.

Solution

Maths sir · 4 steps

Use the tangency condition c² = 25 - 16m² twice: once to find the slopes, once to build Q's tangents.

1. Tangency condition. Put y = mx + c into y²/25 - x²/16 = 1 and set the discriminant to zero. This gives the condition for y = mx + c to touch H.

   c² = 25 - 16m²

2. Slopes from P. The line passes through (4, 1), so c = 1 - 4m. Then (1 - 4m)² = 25 - 16m² gives 4m² - m - 3 = 0, so m = 1 or m = -3/4. That makes |m₁| = 1 and |m₂| = 3/4.

   Slope	c² = 25 - 16m²	c for a positive x-intercept	x-intercept -c/m
   1	9	-3	α = 3
   3/4	16	-4	β = 16/3

3. Locate Q. Q lies on both tangents y = x - 3 and y = 3/4x - 4. Equating them gives x/4 = -1, so Q is (-4, -7). Then PQ² = 8² + 8² = 128 and αβ = 3 × 16/3 = 16.
4. So. (PQ)²/αβ = 128/16 = 8.

The idea to keep. For tangents to a hyperbola, write each tangent as y = mx + c and use the condition on c². With a positive slope, a positive x-intercept forces c to be negative, and that fixes the sign.

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