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If the double ordinate of the parabola $y^2=8 x$ is of length 16 , then the angle subtended by it at the vertex of the parabola is
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Verified Answer
The correct answer is:
$\frac{\pi}{2}$
Let $A\left(x_1, y_1\right)$ and $B\left(x_1,-y_1\right)$ are the co-ordinate of end points of double ordinate length of $A B=2 y_1=16$
$$
A\left(x_1, 8\right) \text { and } B\left(x_1,-8\right)
$$
$A$ and $B$ lies on parabola. So,
$$
64=8 x_1 \Rightarrow x_1=8
$$
So, coordinate of $A$ and $B$ are $(8,8),(8,-8)$
$$
\begin{aligned}
\Rightarrow \quad \tan \alpha & =\frac{8}{8}=1 \\
\alpha & \left.=\frac{\pi}{4} \quad \text { (Angle with } X \text {-axis made by } O A\right)
\end{aligned}
$$
Angle subtended by double ordinate $A B$ at vertex
$$
\begin{aligned}
& =2 \alpha \\
& =2 \cdot \pi / 4=\pi / 2
\end{aligned}
$$
$$
A\left(x_1, 8\right) \text { and } B\left(x_1,-8\right)
$$
$A$ and $B$ lies on parabola. So,
$$
64=8 x_1 \Rightarrow x_1=8
$$
So, coordinate of $A$ and $B$ are $(8,8),(8,-8)$
$$
\begin{aligned}
\Rightarrow \quad \tan \alpha & =\frac{8}{8}=1 \\
\alpha & \left.=\frac{\pi}{4} \quad \text { (Angle with } X \text {-axis made by } O A\right)
\end{aligned}
$$
Angle subtended by double ordinate $A B$ at vertex
$$
\begin{aligned}
& =2 \alpha \\
& =2 \cdot \pi / 4=\pi / 2
\end{aligned}
$$
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