JEE MAIN - Mathematics (2025 - 23rd January Evening Shift - No. 22)
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 $\lambda_2, \lambda_1<\lambda_2$, then $12 \lambda_1+29 \lambda_2$ is equal to ________ .
Answer
15
Explanation
$$y^2=4(x+4)$$
Equation of circle
$$(x+3)^2+y^2=25$$
Passes through the point of intersection of two lines $3 x-y=0$ and $x+\lambda y=4$ which is $\left(\frac{4}{3 \lambda+1}, \frac{12}{3 \lambda+1}\right)$, after solving with circle, we get
$$\begin{aligned} & \lambda=-\frac{7}{6}, 1 \\ & 12 \lambda_1+29 \lambda_2 \\ & -14+29=15 \end{aligned}$$
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