JEE Main · 2025 · 23 Jan 2025 · Shift 2 · Parabola

The focus of the parabola

y^{2} = 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

λ_{1}

and

λ_{2}

λ_{1} < λ_{2}

, then

12 λ_{1} + 29 λ_{2}

is equal to ______.

Answer: 15

Solution

Rewrite

y^{2} = 4 x + 16

y^{2} = 4 (x + 4)

, a parabola with vertex

(- 4, 0)

and

4 a = 4

, so its focus is at

(- 4 + 1, 0) = (- 3, 0)

.

Circle

C

has centre

(- 3, 0)

and radius 5:

(x + 3)^{2} + y^{2} = 25

Intersection of

3 x - y = 0

and

x + λ y = 4

: from

y = 3 x

, substitute to get

x (1 + 3 λ) = 4

, so

x = \frac{4}{1 + 3 λ}, y = \frac{12}{1 + 3 λ}

This point lies on

C

(\frac{4}{1 + 3 λ} + 3)^{2} + (\frac{12}{1 + 3 λ})^{2} = 25

(9 λ + 7)^{2} + 144 = 25 (1 + 3 λ)^{2}

81 λ^{2} + 126 λ + 193 = 225 λ^{2} + 150 λ + 25

144 λ^{2} + 24 λ - 168 = 0 \Rightarrow 6 λ^{2} + λ - 7 = 0

(6 λ + 7) (λ - 1) = 0

λ = 1

λ = - \frac{7}{6}

. With

λ_{1} < λ_{2}

λ_{1} = - \frac{7}{6}

λ_{2} = 1

12 λ_{1} + 29 λ_{2} = 12 (- \frac{7}{6}) + 29 (1) = - 14 + 29 = 15

JEE Main · 2025 · 23 Jan 2025 · Shift 2 · Integer answer