ParabolaHardBloom L3

Question

If the tangents to the parabola $y^{2} = 4ax$ at $\left( x_{1},y_{1} \right)$ and $\left( x_{2},y_{2} \right)$ meet at $\left( x_{3},y_{3} \right)$, then:

Options

A.$x_{1},x_{3},x_{2}$ are in A.P.
B.$x_{1},x_{3},x_{2}$ are in G.P.
C.$y_{1},y_{3},y_{2}$ are in G.P.
D.$y_{1},y_{3},y_{2}$ are in A.P.

Solution

Sol. [Key Concept: If tangents at $t_{1},t_{2}$ meet at $(x_{3},y_{3})$ then $x_{3} = {at}_{1}t_{2}$, $y_{3} = a(t_{1} + t_{2})$; hence $x_{1},x_{3},x_{2}$ are in GP and $y_{1},y_{3},y_{2}$ in AP.]

$${\left( at_{1}^{2} \right)\left( at_{2}^{2} \right) = \left( at_{1}t_{2} \right)^{2} }{\Rightarrow \ x_{1}x_{2} = x_{3}^{2}}$$

$$\frac{2at_{1} + 2at_{2}}{2} = a\left( t_{1} + t_{2} \right)$$

$$\Rightarrow \ \frac{y_{1} + y_{2}}{2} = y_{3} $$

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