1. **Problem statement:** Given the parabola $$y^2=2ax$$, a double ordinate is a chord perpendicular to the axis of the parabola. The length of this double ordinate is given as $$4a$$. We need to prove that the lines joining the vertex to the ends of this double ordinate are at right angles. 2. **Recall the parabola and double ordinate:** The parabola is $$y^2=2ax$$ with vertex at the origin $$(0,0)$$. 3. **Length of double ordinate:** Let the double ordinate intersect the parabola at points $$P(x_1,y_1)$$ and $$Q(x_1,-y_1)$$ since it is perpendicular to the x-axis (axis of parabola). 4. From the parabola equation, $$y_1^2=2ax_1$$. 5. The length of the double ordinate $$PQ=|y_1 - (-y_1)|=2|y_1|=4a$$ (given). 6. Therefore, $$2|y_1|=4a \implies |y_1|=2a$$. 7. Substitute $$y_1=2a$$ into the parabola equation: $$ (2a)^2=2ax_1 \implies 4a^2=2ax_1 \implies x_1=\frac{4a^2}{2a}=2a $$. 8. So the points are $$P(2a,2a)$$ and $$Q(2a,-2a)$$. 9. **Find slopes of lines from vertex to ends:** - Slope of $$VP = \frac{2a-0}{2a-0}=1$$ - Slope of $$VQ = \frac{-2a-0}{2a-0}=-1$$ 10. **Check if lines are perpendicular:** Product of slopes $$=1 \times (-1)=-1$$. 11. Since the product of slopes is $$-1$$, the lines $$VP$$ and $$VQ$$ are perpendicular. **Final answer:** The lines joining the vertex to the ends of the double ordinate are at right angles.