A double ordinate of the parabola A double ordinate of the parabola  $y^2=2ax $ is of length  $4a $. Prove that the lines joining the vertex to its ends are at right angles. 
What does double ordinate actually mean?  Seeing that the length is  $4a $, I guess it is Latus Rectum. How do I proceed this? 
 A: So if $y= \pm 2a$ and $x=2a$ then we have points $A(2a,2a)$, $B(2a,-2a)$ and $C(0,0)$ which is obviously right triangle.
A: If $x=2at^2, y^2=(2at)^2\implies y=\pm2at$
WLOG the two ends be $P(2ap^2,2ap)$ and  $Q(2aq^2,2aq)$
So, the gradient of $PQ$ will be $$\dfrac{2a(q-p)}{2a(q^2-p^2)}=\dfrac1{q+p}$$
But it's perpendicular to the axis $y=0$ of the parabola
$\implies q+p=0\iff q=-p$
Now the length of $PQ$ will be $$\sqrt{(2ap^2-2aq^2)^2+(2ap-2aq)^2}=4a|q|$$
Finally, if $O(0,0)$ is vertex, 
 what will be product of gradients of $OP,OQ$?
A: Double ordinate is perpendicular to axis of parabola, hence let the end points of double ordinate be $$P=(at^2,2at)\mbox{ and }P^{\prime}=(at^2,-2at)$$
$$PP^{\prime}=4a$$
$$\sqrt{(at^2-at^2)^2+(2at+2at)^2}=4a$$
$$\sqrt{(4at)^2}=4a$$
$$4at=4a\implies t=1$$
Hence $P=(at^2,2at)=(a,2a)$ and $P^{\prime}=(at^2,-2at)=(a,-2a)$
Vertex is $O(0,0)$
Slope of $OP=m_1=2$
Slope of $OP^{\prime}=m_2=-\dfrac 12$
$$\implies m_1\times m_2=-1$$
Hence $\angle POP^{\prime}=\dfrac{\pi}{2}$
Therefore, they form right triangles.
