If the point (a, 2a) is an interior point of | vedclass

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(B) Given the parabola $y^2 = 16x$. Comparing this with the standard form $y^2 = 4px$,we get $4p = 16$,so $p = 4$. The focus of the parabola is $S = (

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$0 < a < 4$

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(B) Given the parabola $y^2 = 16x$. Comparing this with the standard form $y^2 = 4px$,we get $4p = 16$,so $p = 4$. The focus of the parabola is $S = (4, 0)$. The double ordinate through the focus is the line $x = 4$. Since the point $P(a, 2a)$ lies in the interior region of the parabola $y^2 - 16x = 0$,we must have $y^2 - 16x Substituting the point $(a, 2a)$ into the inequality: $(2a)^2 - 16a $4a^2 - 16a $4a(a - 4) This implies $0 Additionally,the point $P(a, 2a)$ must lie to the left of the double ordinate $x = 4$ to be within the bounded region. Thus,$a Combining both conditions,we get $0 < a < 4$.

If the point $(a, 2a)$ is an interior point of the region bounded by the parabola $y^2 = 16x$ and the double ordinate through the focus,then

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