If a double ordinate of the parabola y^2 = 4a | vedclass

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(C) The equation of the parabola is $y^2 = 4ax$. The vertex is at $O(0, 0)$.Let the double ordinate be at $x = h$. Then $y^2 = 4ah$,so $y = \pm 2\sqrt

Vedclass · Accepted Answer

$90$

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(C) The equation of the parabola is $y^2 = 4ax$. The vertex is at $O(0, 0)$.Let the double ordinate be at $x = h$. Then $y^2 = 4ah$,so $y = \pm 2\sqrt{ah}$.The length of the double ordinate is $2 	imes 2\sqrt{ah} = 4\sqrt{ah}$.Given the length is $8a$,we have $4\sqrt{ah} = 8a$,which implies $\sqrt{ah} = 2a$,so $ah = 4a^2$,hence $h = 4a$.The ends of the double ordinate are $P(4a, 4a)$ and $Q(4a, -4a)$.The slope of $OP$ is $m_1 = \frac{4a - 0}{4a - 0} = 1$.The slope of $OQ$ is $m_2 = \frac{-4a - 0}{4a - 0} = -1$.Since $m_1 	imes m_2 = 1 	imes (-1) = -1$,the lines $OP$ and $OQ$ are perpendicular to each other.Therefore,the angle between them is $90^\circ$.

If a double ordinate of the parabola $y^2 = 4ax$ has a length of $8a$,then the angle between the lines joining the vertex of the parabola to the ends of this double ordinate is ............... $^\circ$.

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