If m ₁ and m ₂ are the slopes of the lines represented by a x^2+2 ~h x y + by ^2=0 satisfying the condition 16 ~h ^2=25 a ~b , then .

Step-by-step solution

For the homogeneous equation ax^2 + 2hxy + by^2 = 0 with slopes m₁ and m₂ , the standard relations are m₁ + m₂ = - 2h b and m₁ m₂ = a b . Given the condition 16h^2 = 25ab , substitute the expressions 2h = -b(m₁ + m₂) and a = b m₁ m₂ : 16 ( b^2(m₁ + m₂)^2 4 ) = 25b^2 m₁ m₂ Simplifying yields 4(m₁ + m₂)^2 = 25 m₁ m₂ , which expands to: 4m₁^2 - 17m₁ m₂ + 4m₂^2 = 0 Dividing through by m₂^2 gives 4x^2 - 17x + 4 = 0 where x = m₁ m₂ . Solving this quadratic: x = 17 289 - 64 8 = 17 15 8 Thus x = 4 or x = 1 4 , meaning m₁ =