The focus of the parabola y^2=4x+16 is the centre of a circle C of radius 5.

Question

The focus of the parabola $y^2=4x+16$ is the centre of a 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 \_\_\_\_\_\_\_\_.

Answer: 15

Accepted Answer

$y^2=4(x+4)$, focus $(-3,0)$; circle $C:(x+3)^2+y^2=25$.

Intersection of the lines: from $y=3x$, $x(1+3\lambda)=4 \Rightarrow \left(\dfrac{4}{3\lambda+1},\dfrac{12}{3\lambda+1}\right)$.

Substituting into $C$ (let $u=\dfrac{1}{3\lambda+1}$, so the point is $(4u,12u)$):

$$(4u+3)^2+144u^2=25 \;\Rightarrow\; 160u^2+24u-16=0 \;\Rightarrow\; 20u^2+3u-2=0.$$

$$u=\frac14 \text{ or } -\frac25 \;\Rightarrow\; \lambda=1 \text{ or } -\frac76.$$

So $\lambda_1=-\dfrac76,\ \lambda_2=1$:

$$12\lambda_1+29\lambda_2=12\left(-\frac76\right)+29(1)=-14+29=15.$$

Solutions · 1