LOCUS help (1 Viewer)

[bmn]

Stuck on acouple of textbook questions... First time stuck on any in the Maths in Focus book.... This would have to be my weakest topic so far.

1) A parabolic satellite dish has a diameter of 4m at a depth of 0.4m. Find the depth at which its diameter is 3.5m, correct to 1 decimal place.
Very confused with this one...

2) The normal of the parabola x^2 = 18y at (-6, 2) cuts the parabola again at Q. Find the coordinates of Q.
My answer was (-21, 24.5) It works if I sub it back into equation and into the normal of the bit (my equation of the normal was 3x + 2y + 14 = 0, if anywehre I probably went wrong here.) Books answer was different, can some1 explain what I am doing wrong?

3) Find the equation of the tangent at (8, 4) on the parabola x^2 = 16y. This tangent meets the tangent at the vertex of the parabola at point R. Find the coordinates of R.
I got the equation of tangent to be x - y - 4 = 0, which was right. No idea what to do from there (Is it just me or is the question badly worded?)

4)

.............|............
.............|............
.........../.|.\...........
........../..|...\.........
........./...|.6m\....... (hieght)
......../....|.......\.....
___/___|____\__
.......-3m-..-3m-....

badly drawn, but its meant to look like a parabola...

(a) Find the equation of this parabola (I'm probably forgetting something from a earlier chapter)
(b) Find the coordinates of its forucs and the equation of its directrix (Even with the correct answer for (a) I cant seem to find the focal length or any other way?)

 

[cutemouse]

Hahaha, I remember doing that one early this year.

I wouldn't worry about it though, nothing like that came up in my exams for prelim.

Stuck on acouple of textbook questions... First time stuck on any in the Maths in Focus book.... This would have to be my weakest topic so far.

1) A parabolic satellite dish has a diameter of 4m at a depth of 0.4m. Find the depth at which its diameter is 3.5m, correct to 1 decimal place.
Very confused with this one...

 

[Timothy.Siu]

1) A parabolic satellite dish has a diameter of 4m at a depth of 0.4m. Find the depth at which its diameter is 3.5m, correct to 1 decimal place.

y=0.4, x=+-2

x^2=4ay
lhs=4 rhs=1.6a=4
a=4/1.6=5/2

therefore x^2=5/2x4y

when x=+-1.75 y=0.30625m

2) The normal of the parabola x^2 = 18y at (-6, 2) cuts the parabola again at Q. Find the coordinates of Q.
y=x^2/18
y'=x/9
when x=-6 y'=-2/3
equation of normal: y-2=3/2(x+6)
y=1.5x+11
y=x^2/18

1.5x+11=x^2/18
x^2-27x-198=0
x=33 or -6
so, Q(33,60.5)

3) Find the equation of the tangent at (8, 4) on the parabola x^2 = 16y. This tangent meets the tangent at the vertex of the parabola at point R. Find the coordinates of R.

no they question isn't badly worded
y'=x/8 at x=8 y'=1
y-4=x-8
y=x-4
tangent at vertex is y=0 (obvious)
sub into tangent
R(4,0)

 

[Timothy.Siu]

4)

.............|............
.............|............
.........../.|.\...........
........../..|...\.........
........./...|.6m\....... (hieght)
......../....|.......\.....
___/___|____\__
.......-3m-..-3m-....

badly drawn, but its meant to look like a parabola...

(a) Find the equation of this parabola (I'm probably forgetting something from a earlier chapter)
(b) Find the coordinates of its forucs and the equation of its directrix (Even with the correct answer for (a) I cant seem to find the focal length or any other way?)

parabola is y=-2/3(x+3)(x-3)
b)y=-2/3(x^2-9)
-3y=2x^2-18
x^2=-3/2y+9
x^2=-4x3/8(y-6)

focus: (0,45/8)
directrix: y=49/8

 

[bmn]

4)

.............|............
.............|............
.........../.|.\...........
........../..|...\.........
........./...|.6m\....... (hieght)
......../....|.......\.....
___/___|____\__
.......-3m-..-3m-....

badly drawn, but its meant to look like a parabola...

(a) Find the equation of this parabola (I'm probably forgetting something from a earlier chapter)
(b) Find the coordinates of its forucs and the equation of its directrix (Even with the correct answer for (a) I cant seem to find the focal length or any other way?)

parabola is y=-2/3(x+3)(x-3)
b)y=-2/3(x^2-9)
-3y=2x^2-18
x^2=-3/2y+9
x^2=-4x3/8(y-6)

focus: (0,45/8)
directrix: y=49/8

The answer in the book is different.

To part a is 8x^2 + 9y - 72 = 0

and for part b

Focus: (0, 161/32)
directrix y=72/32

 

[Timothy.Siu]

The answer in the book is different.

To part a is 8x^2 + 9y - 72 = 0

and for part b

Focus: (0, 161/32)
directrix y=72/32

well if i got part a wrong, wuda got part b wrong and i'm right, unless u typed the question wrong

 

[cutemouse]

Which EX is it from Math in Focus? I'll try to have a look for you..

 

[blakdevil]

well hi, im new
im currently only in yr 10 and yr 11 09
but then im doin maths accelerated and did my HSC 2 unit recently, tho i think i really screwed up

i find that im weak at locus

 

[Timothy.Siu]

well hi, im new
im currently only in yr 10 and yr 11 09
but then im doin maths accelerated and did my HSC 2 unit recently, tho i think i really screwed up

i find that im weak at locus

nice, dont worry, u still hav 2 more years, plenty of time to get better

 

[bmn]

well if i got part a wrong, wuda got part b wrong and i'm right, unless u typed the question wrong

I did Meant to be 8 instead of 6... now it makes sence...

 

[Timothy.Siu]

I did Meant to be 8 instead of 6... now it makes sence...

lol, thought so