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JEE Main 202523 JanuaryEvening Shift

Exam: JEE Main
Year: 2025
Date: 23 January
Shift: Evening Shift
Subject: Mathematics
Chapter: Parabola

JEE Main 2025 · 23 January · Evening Shift · Mathematics · Parabola

The focus of the parabola y^2=4 x+16 is the centre of the circle C...

The focus of the parabola

y^2=4 x+16

is the centre of the circle

C

of radius 5 . If the values of

\lambda

, for which C passes through the point of intersection of the lines

3 x-y=0

and

x+\lambda y=4

, are

\lambda_1

and

\lambda_2, \lambda_1 \lt \lambda_2

, then

12 \lambda_1+29 \lambda_2

is equal to

\qquad

Correct answer

0

Step-by-step solution

y^2=4(x+4)

Equation of circle

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

Passes through the point of intersection of two lines

3 x-y=0 and x+\lambda y=4 \left(\frac{4}{3 \lambda+1}, \frac{12}{3 \lambda+1}\right) \text{, we get }

\begin{aligned} & \lambda=-\frac{7}{6}, 1 \\ & 12 \lambda_1+29 \lambda_2 \\ & -14+29=15 \end{aligned}

$$

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