Parabola
Parabola
nta_pyq_2025_apr
Grade 11
Question:
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 _____.
Step-by-Step Solution
Key Concept: Rewrite the parabola as $y^2=4(x+4)$ to read off the focus; form the circle equation; substitute the parameterised intersection point of the two lines and solve the resulting quadratic in $\lambda$.
$y^2=4(x+4)$ has focus at $(-3,0)$. Circle $C$: $(x+3)^2+y^2=25$. Intersection of $3x-y=0$ (i.e., $y=3x$) and $x+\lambda y=4$: substituting $y=3x$ gives $x=\tfrac{4}{3\lambda+1}$, $y=\tfrac{12}{3\lambda+1}$. Substituting into the circle equation: $$\left(\tfrac{4}{3\lambda+1}+3\right)^2+\left(\tfrac{12}{3\lambda+1}\right)^2=25.$$ Expanding: $(7+9\lambda)^2+144=25(3\lambda+1)^2$, which simplifies to $6\lambda^2+\lambda-7=0$, giving $\lambda=1$ or $\lambda=-\tfrac{7}{6}$. So $\lambda_1=-\tfrac{7}{6}$, $\lambda_2=1$, and $12(-\tfrac{7}{6})+29(1)=-14+29=15$.
Correct Answer: 15