The triangle PQR 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 QR 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 $\frac{2 A}{a}$
The triangle $P Q R$ of area $A$ is inscribed in parabola $y^{2} = 40 x$ such that vertex $P$ lies at vertex of parabola and base $Q R$ is focal chord. The modulus of difference of ordinates of points $Q$ and $R$ is:
We find $∣ ∣ ∣ 2 a t - (- \frac{2 a}{t}) ∣ ∣ ∣ = ∣ ∣ ∣ 2 a (t + \frac{1}{t}) ∣ ∣ ∣ ⟶ e q n .1$
Area of $△ P Q R = \frac{1}{2} | x_{1} (y_{2} - y_{3} + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}) |$
$= \frac{1}{2} ∣ ∣ ∣ a t^{2} (- \frac{2 a}{t}) + \frac{a}{t^{2}} (- 2 a t) ∣ ∣ ∣$
$= \frac{1}{2} ∣ ∣ ∣ - 2 a^{2} t - \frac{2 a^{2}}{t} ∣ ∣ ∣$
$= ∣ ∣ ∣ a^{2} (t + \frac{1}{t}) ∣ ∣ ∣ = A ⟶ e q n .2$
Put value of $(t + \frac{1}{t})$ from $e q n .2$ in $e q n .1$
$∣ ∣ y_{Q} - y_{R} ∣ ∣ = ∣ ∣ ∣ 2 a (t + \frac{1}{t}) ∣ ∣ ∣ = 2 a (\frac{A}{a^{2}}) = \frac{2 A}{a}$
Answer: $\frac{2 A}{a}$

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