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 (integer)
15
Solution
<p>$y^2=4(x+4)$</p>
<p>Equation of circle</p>
<p>$(x+3)^2+y^2=25$</p>
<p>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</p>
<p>$$\begin{aligned}
& \lambda=-\frac{7}{6}, 1 \\
& 12 \lambda_1+29 \lambda_2 \\
& -14+29=15
\end{aligned}$$</p>
About this question
Subject: Mathematics · Chapter: Conic Sections · Topic: Parabola
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