Question

Consider two parabolas \(P_1,\ P_2\) and a line \(L\):
\[ P_1:\ y=4x^2,\qquad P_2:\ y=x^2+27,\qquad L:\ y=\alpha x \] If the area bounded by \(P_1\) and \(P_2\) is six times the area bounded by \(P_1\) and \(L\), find \(\alpha\).

• A. \(16\)
• B. \(18\)
• C. \(20\)
• D. \(12\)

Hint:

For area comparison problems, first compute both areas symbolically, then apply the given ratio to solve for the parameter.

Answer 1

The Correct Option is D

Step 1: Area between the two parabolas
Points of intersection of \(P_1\) and \(P_2\): \[ 4x^2=x^2+27 \Rightarrow 3x^2=27 \Rightarrow x=\pm 3 \] Area \(A_1\): \[ A_1=\int_{-3}^{3}\bigl[(x^2+27)-4x^2\bigr]dx =\int_{-3}^{3}(27-3x^2)\,dx \] \[ A_1=2\int_{0}^{3}(27-3x^2)\,dx =2\left[27x-x^3\right]_{0}^{3} =2(81-27)=108 \]
Step 2: Area between \(P_1\) and the line \(L\)
Intersection of \(y=4x^2\) and \(y=\alpha x\): \[ 4x^2=\alpha x \Rightarrow x(4x-\alpha)=0 \] So, \[ x=0,\quad x=\frac{\alpha}{4} \] Area \(A_2\): \[ A_2=\int_{0}^{\alpha/4}\bigl(\alpha x-4x^2\bigr)\,dx \] \[ A_2=\left[\frac{\alpha x^2}{2}-\frac{4x^3}{3}\right]_{0}^{\alpha/4} =\frac{\alpha^3}{32}-\frac{\alpha^3}{48} =\frac{\alpha^3}{96} \]
Step 3: Use the given condition
Given: \[ A_1=6A_2 \] \[ 108=6\cdot\frac{\alpha^3}{96} \Rightarrow 108=\frac{\alpha^3}{16} \] \[ \alpha^3=1728 \Rightarrow \alpha=12 \]
Final Answer:
\[ \boxed{12} \]