The focus of the parabola y²=4 x+16 is the centre of the circle C of radius 5 . If the values of λ, for which C passes through the point of intersection of the lines 3 x-y=0 and x+λ y=4, are λ₁ and λ₂, λ₁<λ₂, then 12 λ₁+29 λ₂ is equal to ________ .

Solution

Maths sir · 5 steps

Find the focus first, then write the intersection point in terms of λ and drop it into the circle.

1. Focus of the parabola. Write y² = 4x + 16 as y² = 4(x+4). This is Y² = 4aX with X = x+4 and a = 1, so the vertex is (-4, 0) and the focus is (-4+1, 0) = (-3, 0).
2. Equation of C. With centre (-3, 0) and radius 5, the circle is (x+3)² + y² = 25.
3. Intersection point. From 3x - y = 0, put y = 3x into x + λ y = 4 to get x(1+3λ) = 4. Let t = 1 + 3λ. Then the point is (4/t, 12/t).

   (4/t+3)² + (12/t)² = 25 ⇒ 2t² - 3t - 20 = 0

4. Solve for lambda. Multiplying by t² gives (4+3t)² + 144 = 25t², which simplifies to 16t² - 24t - 160 = 0, that is 2t² - 3t - 20 = 0. So t = 4 or t = -5/2, which gives λ = 1 or λ = -7/6.
5. So. λ₁ = -7/6 and λ₂ = 1, so 12λ₁ + 29λ₂ = -14 + 29 = 15.

The idea to keep. For a shifted parabola, rewrite it as Y² = 4aX before reading off the focus. When a point depends on a parameter, substitute it into the curve and solve for that parameter.

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