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

Question

Dr. Abubacker · Accepted Answer

[Solution by Dr. Abubacker] $y^2=4(x+4)$ has focus $(-3,0)$. The lines meet at $\left(\frac4t,\frac{12}t\right)$ with $t=1+3\lambda$. Requiring distance $5$ from $(-3,0)$: $\left(\frac4t+3\right)^2+\frac{144}{t^2}=25\Rightarrow2t^2-3t-20=0$, so $t=4$ or $-\frac52$. Then $\lambda_2=1$ and $\lambda_1=-\frac76$. $12\lambda_1+29\lambda_2=-14+29=15$. Answer: 15.

Explanation

Reviewed by Dr. Abubacker

Want to attempt the full paper?