JEE Main 2025 (Online) 23rd January Evening ShiftQuestion 1
The focus of the parabola $y^2=4x+16$ 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 $3x-y=0$ 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 ______.
Circle
Answer$15$
Solution
Concept. Write the meeting point of the two lines in terms of $\lambda$ and put it on the circle. Formulas used. (1) Focus of $y^2=4a(x-h)$ is $(h+a,0)$Step 1 - Centre. $y^2=4(x+4)$: $h=-4$, $a=1$, focus $(-3,0)$. Circle: $(x+3)^2+y^2=25$. Step 2 - Meeting point. $y=3x$ in $x+\lambda y=4$: $x(1+3\lambda)=4$, so $x=\frac{4}{1+3\lambda}$, $y=3x$. Step 3 - On the circle. $$(x+3)^2+9x^2=25\Rightarrow 10x^2+6x-16=0\Rightarrow 5x^2+3x-8=0.$$ $(x-1)(5x+8)=0\Rightarrow x=1$ or $x=-\frac85$. Step 4 - $\lambda$. $x=1$: $1+3\lambda=4\Rightarrow\lambda=1$. $x=-\frac85$: $1+3\lambda=-\frac52\Rightarrow\lambda=-\frac76$. Step 5 - Value. $\lambda_1=-\frac76$, $\lambda_2=1$: $12\lambda_1+29\lambda_2=-14+29=15$. Answer: 15