Let a line L₁ be tangent to the hyperbola x²/16 - y²/4 = 1 and let L₂ be the line passing through the origin and perpendicular to L₁. If the locus of the point of intersection of L₁ and L₂ is (x² + y²)² = α x² + β y², then α + β is equal to _____________.

Solution

Maths sir · 4 steps

Write the tangent in slope form, then use the perpendicular from O to swap the slope m for -h/k.

1. Tangent in slope form. Here a² = 16 and b² = 4. Every tangent L₁ can be written as y = mx + c with the tangency condition c² = a²m² - b² = 16m² - 4.
2. Use the perpendicular. Let the intersection point be P(h, k). L₂ joins the origin to P, so its slope is k/h. Since L₁ ⊥ L₂, m = -h/k.
3. Find c. P lies on L₁, so k = mh + c. This gives c = k + h²/k = (h² + k²)/k.

   (h²+k²)²/k² = 16h²/k² - 4 ⇒ (h²+k²)² = 16h² - 4k²

4. So. The locus is (x²+y²)² = 16x² - 4y². So α = 16, β = -4, and α + β = 12.

The common slip: 20. Taking β = +4 by copying the ellipse result a²x² + b²y² gives 20, not 12. For a hyperbola the b² term is negative.

The idea to keep. The foot of the perpendicular from the centre to a tangent of x²/a² - y²/b² = 1 traces (x²+y²)² = a²x² - b²y². Watch the minus sign in front of b².

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