A double ordinate of the curve $y^{2} = 4 p x$ is of length $8 p;$ prove that the lines from the vertex to its two ends are at right angles.

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Solution

Length of latus rectum $= 8 p$
$\Rightarrow$ semi latus rectum $= 4 p$
So the ordinates of end points are $4 p$ and $- 4 p$
At $y = \pm 4 p$
${(\pm 4 p)}^{2} = 4 p x 4 p x = 16 p^{2} \Rightarrow x = 4 p$
So, the coordinates of end points are $L (4 p, 4 p)$ and $L^{'} (4 p, - 4 p)$
Slope of $O L$ $= \frac{4 p - 0}{4 p - 0} = 1$
Slope of $O L^{'}$ $= \frac{4 p - 0}{- 4 p - 0} = - 1$
Slope of $O L$ $\times$ slope of $O L^{'}$ $= 1 \times - 1 = - 1$
Product of slopes of two lines is $- 1$ .Hence they are perpendicular

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A double ordinate of the curve y2=4px is of length 8p; prove that the lines from the vertex to its two ends are at right angles.

A double ordinate of the curve $y^{2} = 4 p x$ is of length $8 p;$ prove that the lines from the vertex to its two ends are at right angles.