<p>Let equation of line passing through <em>P</em>(1, <em>t</em>) be
\(\frac{x-1}{\cos\theta} = \frac{y-t}{\sin\theta} = r\). Line meets the parabola at <em>A</em> and <em>B</em>. Find the maximum value of <em>t</em> such that
\[\frac{|t^2-4|}{3|t|} = \sin^2\theta \leq 1\]
leads to \(|t| \leq 4\). What is the maximum value of <em>t</em>?</p>
Step-by-Step Solution
Key Concept: The constraint $\sin^2\theta \leq 1$ combined with $\frac{|t^2-4|}{3|t|} = \sin^2\theta$ requires $\frac{|t^2-4|}{3|t|} \leq 1$, which when solved gives the range of valid $t$ values. The maximum occurs at the boundary.
<p><strong>Step 1:</strong> From the given condition, we have $\frac{|t^2-4|}{3|t|} = \sin^2\theta$ where $\sin^2\theta \leq 1$.</p><p><strong>Step 2:</strong> This means $\frac{|t^2-4|}{3|t|} \leq 1$, which gives $|t^2-4| \leq 3|t|$.</p><p><strong>Step 3:</strong> Consider two cases:</p><p><strong>Case 1 (t > 0):</strong> $|t^2-4| \leq 3t$</p><p>If $t^2 \geq 4$ (i.e., $t \geq 2$): $t^2-4 \leq 3t \Rightarrow t^2-3t-4 \leq 0 \Rightarrow (t-4)(t+1) \leq 0$. For $t > 0$: $2 \leq t \leq 4$</p><p>If $t^2 < 4$ (i.e., $0 < t < 2$): $4-t^2 \leq 3t \Rightarrow t^2+3t-4 \geq 0 \Rightarrow (t+4)(t-1) \geq 0$. For $0 < t < 2$: $1 \leq t < 2$</p><p>Combined for $t > 0$: $1 \leq t \leq 4$</p><p><strong>Case 2 (t < 0):</strong> By similar analysis, $-4 \leq t \leq -1$</p><p><strong>Step 4:</strong> The maximum value of $t$ satisfying the constraint is $t = 4$.</p><p>∴ Answer: 4</p>
Correct Answer: 4