If the normal at point $t_1$ on the parabola $y^2 = 4ax$ meets the parabola again at $t_2$, then the relation between $t_1$ and $t_2$ is:

Subject: Mathematics | Topic: Conic Sections | Difficulty: HARD

Options

1. (A) $t_1 t_2 = -1$
2. (B) $t_1 + t_2 = 0$
3. (C) $t_1^2 + t_1 t_2 + 2 = 0$
4. (D) $t_1 t_2 = 4$

Verified Answer

Option (C): $t_1^2 + t_1 t_2 + 2 = 0$

10-Second Shortcut Trick

⚡ 10-Second Exam Shortcut: Eliminate options ending in incorrect units or calculate unit digits first before solving full arithmetic.

Step-by-Step Derivation

1. Recall the property of normals meeting on the parabola $y^2 = 4ax$. If the normal at $t_1$ meets the parabola at $t_3$, then $t_3 = -t_1 - \frac{2}{t_1}$.
2. If the normal at $t_2$ also meets the parabola at the same point $t_3$, equate the expressions for $t_3$ corresponding to $t_1$ and $t_2$.
3. This leads to the fundamental relation $t_1 t_2 = 2$ (or $t_2 = -t_1 - 2/t_1$, simplifying to $t_1^2 + t_1 t_2 + t_2^2$ relationships depending on formulation; specifically $t_1 t_2 = 2$ or $t_1^2 + t_1 t_2 + t_2^2 + 2 = 0$). Wait, standard normal meeting condition on the curve is $t_2 = -t_1 - \frac{2}{t_1}$, giving $t_1^2 + t_1 t_2 + 2 = 0$.

Common Trap to Avoid

⚠️ Common Trap: Misreading inverse proportions or rushing calculation in the last algebraic step.

Exam Tip

💡 Exam Tip: Always verify if question asks for perimeter, area, or remaining fraction.

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