A double ordinate of the parabola

Question

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?

Answer 1

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.

Answer 2

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 $OP,OQ$?

Answer 3

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.