The triangle PQR of area A is inscribed in th | vedclass

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(C) Let the coordinates of $R$ be $(at^2, 2at)$. Since $QR$ is a focal chord,the coordinates of $Q$ are $(\frac{a}{t^2}, -\frac{2a}{t})$.The differenc

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$\frac{2A}{a}$

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(C) Let the coordinates of $R$ be $(at^2, 2at)$. Since $QR$ is a focal chord,the coordinates of $Q$ are $(\frac{a}{t^2}, -\frac{2a}{t})$.The difference of the ordinates is $y_R - y_Q = 2at - (-\frac{2a}{t}) = 2a(t + \frac{1}{t})$.The modulus of this difference is $d = 2a|t + \frac{1}{t}|$.The area $A$ of triangle $PQR$ with vertices $P(0,0)$,$R(at^2, 2at)$,and $Q(\frac{a}{t^2}, -\frac{2a}{t})$ is given by:$A = \frac{1}{2} |x_P(y_R - y_Q) + x_R(y_Q - y_P) + x_Q(y_P - y_R)|$$A = \frac{1}{2} |0 + at^2(-\frac{2a}{t} - 0) + \frac{a}{t^2}(0 - 2at)|$$A = \frac{1}{2} |-2a^2t - \frac{2a^2}{t}| = a^2|t + \frac{1}{t}|$.From this,$|t + \frac{1}{t}| = \frac{A}{a^2}$.Substituting this into the expression for $d$:$d = 2a(\frac{A}{a^2}) = \frac{2A}{a}$.

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:

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