ParabolaHardBloom L3

Question

The triangle $PQR$ of area ' $A$ ' is inscribed in the parabola $y^{2} = 4ax$ such that the vertex $P$ lies at the vertex of the parabola and the base $QR$ is a focal chord. The modulus of the difference of the ordinates of the points $Q$ and $R$ is :

Options

A.$\frac{A}{2a}$
B.$\frac{A}{a}$
C.$\frac{2A}{a}$
D.$\frac{4A}{a}$

Solution

Sol. [Key Concept: With the apex at the vertex and base the focal chord $QR$, the area relates linearly to ${|y}_{Q} - y_{R}|$ via the chord's distance from the vertex.]

Area $=$ modulus of $\frac{1}{2}\left| \begin{matrix} 0 & 0 & 1 \\ at_{1}^{2} & 2at_{1} & 1 \\ at_{2}^{2} & 2at_{2} & 1 \end{matrix} \right|$

$$\begin{matrix} & \ = \left| a^{2}t_{1}t_{2}\left( t_{1} - t_{2} \right) \right| \\ A & \ = \left| - a^{2}\left( t_{1} - t_{2} \right) \right| \end{matrix}$$

$$\left| 2a\left( t_{1} - t_{2} \right) \right| = \frac{2A}{a}$$

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