Let mx+ny+p=0 be the equation of a chord. Now, as pointed out by Judasrising, the homogeneousing technique results in a pair of straight lines, with origin as their point of intersection and this pair passes through the point of intersection of chord and the conic section. 

Lets now look at the justification part:

- The resulting equation is homogeneous. So, (0,0) is a point on the resulting homogeneous equation.

- Let (x1,y1) and (x2,y2) be the points of intersection of the chord and the conic section. It can be easily seen that these two points also satisfy the homogeneous equation. 

So, now we conclude that the resulted homogeneous equation is the equation of a curve, passing through origin and 2 points of intersection of the chord and the conic section. The only which we need to prove is that the equation represents a pair of straight lines.

Try getting a condition for the homogeneous equation to be pair of straight lines. the condition we get will be same as the condition for the chord to intersect the conic section. So, the homogeneous equation represents a pair of straight lines, when ever the given straight line intersects with the conic section.

Hence justified.