If the double ordinate of the parabola y^2 = | vedclass

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(A) Let the coordinates of the end points of the double ordinate be $A(x_1, y_1)$ and $B(x_1, -y_1)$.Given the length of the double ordinate $AB = 2y_

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$\frac{\pi}{2}$

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(A) Let the coordinates of the end points of the double ordinate be $A(x_1, y_1)$ and $B(x_1, -y_1)$.Given the length of the double ordinate $AB = 2y_1 = 16$,so $y_1 = 8$.Thus,the coordinates are $A(x_1, 8)$ and $B(x_1, -8)$.Since $A$ and $B$ lie on the parabola $y^2 = 8x$,we have $8^2 = 8x_1$,which gives $64 = 8x_1$,so $x_1 = 8$.The coordinates of $A$ and $B$ are $(8, 8)$ and $(8, -8)$.Let $\alpha$ be the angle made by $OA$ with the $X$-axis,where $O$ is the vertex $(0, 0)$.Then $	an \alpha = \frac{8}{8} = 1$,which implies $\alpha = \frac{\pi}{4}$.The angle subtended by the double ordinate $AB$ at the vertex is $2\alpha = 2 	imes \frac{\pi}{4} = \frac{\pi}{2}$.

If the double ordinate of the parabola $y^2 = 8x$ is of length $16$,then the angle subtended by it at the vertex of the parabola is

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