Parabola
Parabola
nta_pyq_2025_apr
Grade 11
Question:
If the line $3x - 2y + 12 = 0$ intersects the parabola $4y = 3x^2$ at the points $A$ and $B$, then at the vertex of the parabola, the line segment $AB$ subtends an angle equal to
$\tan^{-1}\!\left(\tfrac{4}{5}\right)$
$\tan^{-1}\!\left(\tfrac{9}{7}\right)$
$\tan^{-1}\!\left(\tfrac{11}{9}\right)$
$\tfrac{\pi}{2} - \tan^{-1}\!\left(\tfrac{3}{2}\right)$
Step-by-Step Solution
Key Concept: Find $A$ and $B$ by solving the line and parabola simultaneously; compute slopes $m_{OA}$ and $m_{OB}$ from the vertex $O=(0,0)$; use the angle formula $\tan\theta = |\tfrac{m_1-m_2}{1+m_1m_2}|$.
Substituting $y=\tfrac{3x^2}{4}$ into $3x-2y+12=0$: $3x-\tfrac{3x^2}{2}+12=0 \Rightarrow x^2-2x-8=0 \Rightarrow x=-2,4$. So $A=(-2,3)$ and $B=(4,12)$. Vertex $O=(0,0)$. Slopes: $m_{OA}=-\tfrac{3}{2}$, $m_{OB}=3$. $$\tan\theta=\left|\frac{-\tfrac{3}{2}-3}{1+(-\tfrac{3}{2})(3)}\right|=\left|\frac{-\tfrac{9}{2}}{-\tfrac{7}{2}}\right|=\frac{9}{7}.$$ Hence $\theta=\tan^{-1}\!\left(\tfrac{9}{7}\right)$.
Correct Answer: 2