I also need help with this one if anyone has a quick and easy way of solving them please. Thank you.
(a)
x = 2t - 1 ⇔ x + 1 = 2t ⇒ (x + 1) / 2 = t (1)
Sub this into the second equation for t and we have
y = 4[ (x + 1) / 2 ] ^2 - 3 [( x + 1 ) / 2) + 2 simplify
y = (x + 1)^2 - (3/2)(x + 1) + 2
y = x^2 + 2x + 1 - (3/2)x - 3/2 + 2
y = x^2 + (1/2)x + 3/2 (1)
(b) x coordinate of the vertex = [ -1/2]/ [2 * 1] = -1/4
y coordinate of the vertex = (-1/4)^2 + (1/2)(-1/4) + 3/2 = 23/16
Rearrange (1) as
y - 3/2 = x^2 + (1/2)x complete the square on x
y - 3/2 + 1/16 = x^2 + (1/2)x + 1/16
(y - 23/16) = ( x + 1/4)^2
The vertex = ( -1/4, 23/16)
We have that 4a = 1 so a = 1/4
The focus = ( 1/4 , 23/16 + 1/4) = (- 1/4, 27/16)
The directrix is y = 23/16 - 1/4 = 19/16
(c) When t = -1 we have the point (-3, 9)
The slope of the tangent line at any point x = 2x + 1/2
So...at x = -3, the slope is 2(-3) + 1/2 = -11/2
So....the equation of the tangent line at (-3, 9) is
y = (-11/2) ( x + 3) + 9
y = (-11/2)x - 33/2 + 9
y = (-11/2)x -15/2
See the graph here : https://www.desmos.com/calculator/ukgeugo1pv