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Q. If a double ordinate of the parabola $y^{2}=4ax$ is of length $8a$ , then the triangle formed by double ordinate and line joining vertex with end points of double ordinate is a

NTA AbhyasNTA Abhyas 2022

Solution:

Here, the given parabola $y^{2}=4ax$ , then $\left(a , 4 a\right)$ lies on $y^{2}=4ax$
Solution
$\Rightarrow 16a^{2}=4a\alpha $
$\Rightarrow \alpha =4a$
Slope of $OA=\frac{4 a - 0}{4 a - 0}=1=m_{1}$
Slope of $OA'=\frac{- 4 a - 0}{4 a - 0}=-1=m_{2}$
$m_1 m_2=-1, O A \perp O A^{\prime}$
Hence, $\triangle OAA'$ is a right angled triangle.