Question

A chord is drawn from a point P(1,t) to the parabola y² = 4x which cuts the parabola at A and B. If PA.PB = 3t, then the maximum value of t is equal to

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (D)

4

Answer 2

The product of distances from P(1,t) to the intersection points A and B on the parabola $y^2 = 4x$ is given by $PA \cdot PB = |t^2 - 4| \left(1 + \frac{1}{m^2}\right)$, where $m$ is the slope of the chord. Given $PA \cdot PB = 3t$. So, $|t^2 - 4| \left(1 + \frac{1}{m^2}\right) = 3t$. Since $m$ is real and non-zero, $1 + \frac{1}{m^2} > 1$. Thus, $\frac{3t}{|t^2 - 4|} = 1 + \frac{1}{m^2} > 1$. This implies $\frac{3t}{|t^2 - 4|} > 1$. Also, $PA \cdot PB \ge 0$, so $3t \ge 0$, which means $t \ge 0$. For $t \ge 0$, we analyze the inequality: Case 1: $t^2 - 4 > 0 \implies t > 2$. $3t > t^2 - 4 \implies t^2 - 3t - 4 < 0 \implies (t-4)(t+1) < 0$. For $t>0$, this gives $0 < t < 4$. Combined with $t>2$, we get $2 < t < 4$. Case 2: $t^2 - 4 < 0 \implies 0 \le t < 2$. $3t > -(t^2 - 4) \implies t^2 + 3t - 4 > 0 \implies (t+4)(t-1) > 0$. For $t \ge 0$, this gives $t > 1$. Combined with $0 \le t < 2$, we get $1 < t < 2$. Case 3: $t^2 - 4 = 0 \implies t = 2$ (since $t \ge 0$). If $t=2$, P is $(1,2)$, which is on the parabola. A chord through P has P as one endpoint, so PA=0, $PA \cdot PB = 0$. But $3t = 3(2) = 6$. $0 \ne 6$, so $t \ne 2$. Combining the valid ranges for $t$: $t \in [0, 4)$. The supremum of this set is 4. The question asks for the maximum value, which implies the supremum in this context.