The angle made by a double ordinate of length 8a at the vertex of the parabola $y^{2} = 4 a x$ is

$\frac{π}{3}$

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$\frac{π}{2}$

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$\frac{π}{4}$

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$\frac{π}{6}$

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Solution

The correct option is B
$\frac{π}{2}$

Let PQ be a double ordinate of length 8a.
Then PR = RQ = 4a.
Coordinates of P and Q are (OR, 4a) and (OR, -4a) respectively.
Since P lies on the parabola $y^{2} = 4 a x$ , therefore $(4 a)^{2} = 4 a (O R) \Rightarrow O R = 4 a .$

Thus, the coordinates of P and Q are (4a, 4a) and (4a, -4a) respectively.
Now, $m_{1} = S l o p e o f O P = \frac{4 a - 0}{4 a - 0} = 1 a n d m_{2} = S l o p e o f O Q = \frac{- 4 a - 0}{4 a - 0} = - 1$
Clearly, $m_{1} m_{2} = - 1$
Thus, PQ makes a right angle at the vertex of the parabola.

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