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# I need help with this please.

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2 I also need help with this one if anyone has a quick and easy way of solving them please. Thank you.

Apr 19, 2019

#1
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(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   Apr 19, 2019
#2
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You are the best, thank you so much.

Guest Apr 20, 2019