Geom. Parabola + Locus HELP (1 Viewer)

[bored.of.u]

Hey guys doing locus + geom. of parabola and having trouble with this questions. I got an answer for it but not sure if its correct and i didnt use the geom. of parabola way to solve them, so can u guys show me how to do them using the geom. of parabola way:

Q: Find the eqn of the parabola with vertex (-4, 2) axis parallel to the x axis and has the line x = 6 - 4y as a tangent.

My soln:
m(tangent) = -1/4
General form of the parabola: (y-k)^2 = -4a(x-h)
.: (y-2)^2 = -4a(x+4)
d/dx [(y-2)^2] = -4a
-1/4 = -4a
.: a = 1/16
At a = 1/16 the gradient of the parabola is -1/4
Eqn of the parabola: (y-2)^2 = -1/4.(x+4)
Eqn of the parabola: 4(y-2)^2 = -x-4

Help would be appreciated thanks =D

 

[untouchablecuz]

Hey guys doing locus + geom. of parabola and having trouble with this questions. I got an answer for it but not sure if its correct and i didnt use the geom. of parabola way to solve them, so can u guys show me how to do them using the geom. of parabola way:

Q: Find the eqn of the parabola with vertex (-4, 2) axis parallel to the x axis and has the line x = 6 - 4y as a tangent.

My soln:
m(tangent) = -1/4
General form of the parabola: (y-k)^2 = -4a(x-h)
.: (y-2)^2 = -4a(x+4)
d/dx [(y-2)^2] = -4a
-1/4 = -4a
.: a = 1/16
At a = 1/16 the gradient of the parabola is -1/4
Eqn of the parabola: (y-2)^2 = -1/4.(x+4)
Eqn of the parabola: 4(y-2)^2 = -x-4

Help would be appreciated thanks =D

if the axis is parallel to the x axis it is in the form:

(y-k)^2=-4a(x-h) (the negative sign is obvious if you draw the tangent and the vertex)

k=2 and h=-4

i.e (y-2)^2=-4a(x+4) [1]

x=6-4y [2]

sub [2] in [1]

find the discriminant

discriminant = 0 since the line is tangent (only 1 point of contact)

you end up with

a(8a-1)=0

a=1/8, since a>0

hence the equation is i.e (y-2)^2=-(1/2)(x+4)