chord is drawn through the focus of the parabolay2=6x such that its distance from the vertex ofthis parabola is 52then its slope can be:

Let the equation of the chord be.y−0=mx−32⇒y=mx-3m2 Its distance from the vertex (0, 0) is ca2+b2= m−321+m2=523m1+m2=5squaring on both sides9m2=51+m2 4m2=5 m=±52.