Locus qn. (1 Viewer)

[Getteral09]

hey ppl,Can anyone tell me how to find k in this equation, (x-h)^2=-4a(y-k) of this question: "A parabola has directrix y=5 and focus (-3,3). Find its equation.
Thanks, Getteral

 

[skinnier]

Im pretty sure the answer is

(x+3)^2=-2(y-3)^2

 

[Getteral09]

When i worked that equation out: (x+3)^2=-2(y-3)^2, i found out that answer doesn't equal to the answer in the book which is x^2+6x+4y-7=0.

So how do you work it out?

 

[ThuanSUX]

If the focus is F(-3,3) and directrix is y=5, then your vertex is (-3,4). Same x-co-ord as F and average of y-co-ord. Thus your focal length is 1 (y distance from either F or directrix to vertex).

(x+3)^2=-4(1)(y-4)

 

[Getteral09]

hey, can anyone explain to me on how to do this question?A point is equdistant from the x-axis and y-axis. Find the equation of its locus.
Thanks heaps!

 

[Getteral09]

Can anyone help me on this question?i can't seem to get the answer.Find the equation of the normal to the parabola x^2=4y at the point (-8, 16).
thanks

 

[airie]

Can anyone help me on this question?i can't seem to get the answer.Find the equation of the normal to the parabola x^2=4y at he point (-8, 16).
thanks

You could put it into the parametric form (2at, at²) and so the point (-8, 16) is the point where t=-4 (as a=1 from the equation x²=4y). So the gradient of the tangent at the point is its parameter ie. -4, which means that of the normal is 1/4. Then use the point-gradient form, the equation of the normal is y=x/4+18.

 

[rocky1989]

y=(x^2)/4
y`=x/2
sub in x=-8 for the gradient of the tangent
M of Tangent=-4
then use (Tangent gradient)*(Normal Gradient)=-1

so m=1/4

then use point gradient formula

y-16=0.25(x+8)
x-4y+72=0

 

[Getteral09]

Another question i need help on:The curve y=x^3-2 meets the y-axis at Q. Find the equation of PQ. Do i need to draw a graph to work it out?
Thanks,Getteral

 

[rocky1989]

it doesn't really matter whether you draw the graph or not because the equation y=x^-2 doesn't meet the y-axis....ever (because it is an asymptote)
you have either written the question wrong or this is a horrible question

 

[Getteral09]

Oh, sorry. My bad. A typo. it meant to be:

The curve y=x^3-2 meets the y-axis at Q. Find the equation of PQ. Do i need to draw a graph to work it out?

Thank for attempting a non existing question, though.>.<

 

[SoulSearcher]

Umm, what is P supposed to be?

y = x³ - 2
At the y-axis, x = 0
Therefore y = -2

 

[Getteral09]

but the answers says it is x-y-2=0. Why is that?

 

[SoulSearcher]

Well what is P, for starters? Is it a point?

 

[Riviet]

It definitely would help to draw yourself a diagram. You also haven't told us where P is!