Question ID: #345
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 $\lambda_{2}$, with $\lambda_{1}<\lambda_{2},$ then $12\lambda_{1}+29\lambda_{2}$ is equal to:
Solution:
First, find the focus of the parabola $y^2 = 4(x+4)$.
Let $X = x+4, Y = y$. The parabola is $Y^2 = 4X$.
Focus is at $X = 1, Y = 0 \Rightarrow x+4=1, y=0 \Rightarrow x=-3, y=0$.
So, the focus is $(-3, 0)$.
The circle $C$ has center $(-3, 0)$ and radius 5.
Equation of circle $C$:
$$(x+3)^2 + y^2 = 25$$
Now find the intersection point $P$ of the lines:
1) $y = 3x$
2) $x + \lambda y = 4$
Substitute (1) into (2):
$$x + \lambda(3x) = 4 \Rightarrow x(1+3\lambda) = 4 \Rightarrow x = \frac{4}{1+3\lambda}$$
$$y = 3x = \frac{12}{1+3\lambda}$$
So, $P = \left( \frac{4}{1+3\lambda}, \frac{12}{1+3\lambda} \right)$.
Since circle $C$ passes through $P$, point $P$ satisfies the circle equation:
$$\left(\frac{4}{1+3\lambda} + 3\right)^2 + \left(\frac{12}{1+3\lambda}\right)^2 = 25$$
$$\left(\frac{4 + 3(1+3\lambda)}{1+3\lambda}\right)^2 + \frac{144}{(1+3\lambda)^2} = 25$$
$$\frac{(9\lambda + 7)^2 + 144}{(1+3\lambda)^2} = 25$$
$$(9\lambda + 7)^2 + 144 = 25(1+3\lambda)^2$$
$$81\lambda^2 + 126\lambda + 49 + 144 = 25(1 + 6\lambda + 9\lambda^2)$$
$$81\lambda^2 + 126\lambda + 193 = 25 + 150\lambda + 225\lambda^2$$
$$144\lambda^2 + 24\lambda – 168 = 0$$
Divide by 24:
$$6\lambda^2 + \lambda – 7 = 0$$
$$(6\lambda + 7)(\lambda – 1) = 0$$
Values are $\lambda = 1, -\frac{7}{6}$.
Given $\lambda_1 < \lambda_2$: $$\lambda_1 = -\frac{7}{6}, \quad \lambda_2 = 1$$
We need to compute $12\lambda_1 + 29\lambda_2$:
$$12\left(-\frac{7}{6}\right) + 29(1) = 2(-7) + 29 = -14 + 29 = 15$$
Ans. (15)
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