I assume that "Th,3" is translation by "h" in the x direction and "3" in the y direction- that is, that Th,3(x, y)= (x+h, y+3)- and that "T-2, k" is translation by "-2" in the x direction and "k" in the y direction- that is, that T-2,k(x,y)= (x-2, y+ k).

If that is so then Th,3 x T-2,k(-3, 0)= Th,3(-3-2, 0+k)= Th,3(-5,k)= (-5+h, 3+k)= (-4,8). That is, -5+h= -4 and 3+ k= 8. Now, what are h and k? Once you know that, finding Th,3x T-2,k(2, -1) should be easy.

For a parabola with a y=... directrix, it is of the form:

(x - h)^2 = 4p(y - k)

with vertex (h, k), focus (h, k + p) and directrix y = k - p

With a focus of (3, 6) and a directrix of y = 4, this means:

(h, k + p) = (3, 6) → k + p = 6

y = k - p = 4

→ k = 5, p = 1 (solving the simultaneous equations)

→ vertex is (3, 5)

→ parabola is (x - 3)^2 = 4(y - 5)

which can be rearranged into y = 1/4 x^2 - 3/2 x + 29/4

Oh! i learnt about 8 different types of parabola! 1] y2 = 4ax

2] y2 = -4ax

3] x2 = 4ay

4] x2 = -4ay

5] (y-k)2 = 4a(x-h)

6) (y-k)2 = -4a(x-h)

7] (x-h)2 = 4a(y-k)

8] (x-h)2 = -4a(y-k)

Using the distance formula it is the square root of: (h-x)^2 +(k-y)^2

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