If the chord joining the points P_{1}(x_{1}, | vedclass

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(A) For the parabola $y^{2} = 4ax$, where $4a = 12$, we have $a = 3$. The points on the parabola are $P_{1}(3t_{1}^{2}, 6t_{1})$ and $P_{2}(3t_{2}^{2}

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$288$

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(A) For the parabola $y^{2} = 4ax$, where $4a = 12$, we have $a = 3$. The points on the parabola are $P_{1}(3t_{1}^{2}, 6t_{1})$ and $P_{2}(3t_{2}^{2}, 6t_{2})$.Since the chord $P_{1}P_{2}$ subtends a right angle at the vertex $(0, 0)$, the product of the slopes of $OP_{1}$ and $OP_{2}$ is $-1$.Slope $m_{1} = \frac{6t_{1}}{3t_{1}^{2}} = \frac{2}{t_{1}}$ and $m_{2} = \frac{6t_{2}}{3t_{2}^{2}} = \frac{2}{t_{2}}$.Thus, $(\frac{2}{t_{1}})(\frac{2}{t_{2}}) = -1 \implies t_{1}t_{2} = -4$.Now, $x_{1}x_{2} = (3t_{1}^{2})(3t_{2}^{2}) = 9(t_{1}t_{2})^{2} = 9(-4)^{2} = 9(16) = 144$.And $y_{1}y_{2} = (6t_{1})(6t_{2}) = 36(t_{1}t_{2}) = 36(-4) = -144$.Therefore, $x_{1}x_{2} - y_{1}y_{2} = 144 - (-144) = 144 + 144 = 288$.

If the chord joining the points $P_{1}(x_{1}, y_{1})$ and $P_{2}(x_{2}, y_{2})$ on the parabola $y^{2} = 12x$ subtends a right angle at the vertex of the parabola, then $x_{1}x_{2} - y_{1}y_{2}$ is equal to

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