The Student Room Group
It should just use the normal equation (a) together with the centre (b). So the centre must lie on the normal which represents a radius (well two radii) which is the first line of the ms to t^2=3. Then sub that into the (parametric) equation of the parabola to get the two points in terms of a.

Or it uses the fact that the distance (squared) of a point on the curve fo the centre willl be minimised when the curve touches the circle (at the two points). So the derivative has a stationary point at t^2-3=0 and the same substitution.

Or ...

Note. Id have sketched the parabola and a typical circle. That way its easier to see how the normals can be used as they pass through the centre or how the curve touches the circle at the two points or ... The aim is to get the value of t (or t^2 due to symmetry) which is subbed into the parametric equation of the curve to get the two points.
(edited 1 year ago)
So the marking scheme basically shows a bunch of possible methods.
I'll just briefly explain what the first one means, and if you're curious about others, then just ask.

The first one uses the idea that "the radius of the circle is perpendicular to the tangent at the intersection point".
What else is perpendicular to the tangent? The normal of course. Thus use part (a).
Reply 3
Thank you very much 🙂. Got it now