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

\frac{A}{2 a}

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\frac{A}{a}

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\frac{2 A}{a}

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\frac{4 A}{a}

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Solution

The correct option is C $\frac{2 A}{a}$
Since, $Q R$ is a focal chord
So, vertex of $Q$ is $(a t_{1}^{2}, 2 a t_{1})$ and of $R$ is $(a t_{2}^{2}, 2 a t_{2})$
Area of $Δ P Q R = \frac{1}{2} ∣ ∣ ∣ ∣ ∣ \begin{matrix} 0 & 0 & 1 a t_{1}^{2} & 2 a t_{1} & 1 a t_{2}^{2} & 2 a t_{2} & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣$
$A = \frac{1}{2} | 2 a^{2} t_{1}^{2} t_{2} - 2 a^{2} t_{1} t_{2}^{2} |$
$A = \frac{a}{2} | 2 a t_{1} - 2 a t_{2} |$ ...... $[∵ t_{1} t_{2} = - 1]$
$∴ | 2 a t_{1} - 2 a t_{2} | = \frac{2 A}{a}$

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