Help with this further maths question

B) is correct.
For c) find the focus of the parabola and then connect it to the point of intersection (3,4) with a line. The midpoint of this line segment is the centre of the locus they want you to describe. The two points which are the y intercepts of the lines lie in the circumference of the locus. So write down the equation of the locus.

Is the focus (1,0)?

How do you get that?

I used the fact that the parametric equations for a parabola are x=at^2 and y=2at then I used the y=2t given and solved to find a. Don't know if it's right though.

You want to write it like
https://www.varsitytutors.com/hotmath/hotmath_help/topics/focus-of-a-parabola
Although for your problem, the x and y are flipped. If you've not done it, sketch the parabola and lines and put the focus on there.
To work out the focus properly, you need to eliminate t. Can you show this?

Sorry, but I don't get it. Can you guide me through it a bit more? Thanks.

Eliminate t from the parabola. You can do that.

From y^2=4ax? Do I just substitute x and y into that?

Where does "a" come from? The equation. Is
x = 1/4y^2
From that you can read off the a,h,k in the equation of the parabola. Therefore the focus is ...

(0, 1/4)?

I got a=1/4, h=k=0. So the focus is at
(k+1/(4a), 0) = (1,0)

That’s what I got before!

Sure. I've been trying to make sure you understand why.

I see. I didn't learn the vertex form for a parabola, so that's probably why it was confusing for me. What's the next step? Should I find an equation of a line using the focus and intersection coordinates?

Ok. The midpoint of the line segment joining the two points is simply the midpoint (average) of the two points in the x and y directions . You should just write the point down?
Then the locus is the set of all points which are equidistant from this point. It passes through the y intercepts of the two lines. Again, it should be fairly easy to reason what shape it is and what its parameters are

Ok I think I've almost got it now. So it's a circle equation? Edit: With equation (x-2)^2+(y-2)^2=5?

Sure, it helps to draw the picture.