Locus/Parabola (1 Viewer)

[-pari-]

a parabolic satellite dish has a diametre of 4m at a depth of 0.4m.
find the depth at which its diametre is 3.5m, correct to 1dp.


help on how to go about doing this?

 

[Riviet]

Let the parabola be x²=4ay.
The x-coordinate at any part of the parabola corresponds to half the diameter (ie the radius)
.'. When diameter=4, x=2 and depth (y)=0.4 (or 2/5)
Hence substituting into parabola x²=4ay,
2²=4a(2/5)
a=5/2

.'. equation of parabola is x²= 4.(5/2).y =10y

When diameter=3.5 (or 7/2) , ie when radius (x) = 7/4,
10y=(7/4)²
y=0.3

.'. depth is 0.3 m when diameter is 3.5 m.

 

[-pari-]

oh i get it!


thanks

 

[Dio-Rama]

i dont

y does a=5/2?

y not 2/5?

 

[airie]

Because a*2/5=1, after subbing in the x and y values into the general formula for a parabola. Divide both sides by 2/5 you get a=5/2.

 

[-pari-]

*not related to the above question but still on locus/parabola: didn't want to bother ppl with a whole new thread just on this:

do 2u maths students need to know latus rectum? i've heard they don't? can anyone clarify this for me?

 

[SoulSearcher]

Didn't even know what a latus rectum was until a couple of weeks ago but I'm not sure if you need it for 2 unit, you might want to check the syllabus there.

 

[-pari-]

good idea

also ..this was really stupid of me but i didn't check the answer when i read riviet's reply

redoing the question now...the actual answer is 0.3m :S

i think x^2 = 4ay
is only used when the vertex is at the origin.

 

[airie]

^Actually, when the vertex of a parabola is not at the origin, you could always rotate and translate it until the vertex is

In this case the value of a as Riviet has found is correct. He just made a mistake when he subbed in x=7/4 into the formula. It should have been (7/4)²=10y, therefore y=49/160 which is 0.3 to 1 decimal place

 

[Riviet]

^ Fixed.

Thanks for spotting that airie

 

[-pari-]

awesome. i knew how to do it when i attempted the same question this time round !


um, i have some more questions though, and any help would be really appreciated....

------
Q6 (its part (e) that i couldn't do, but here's all the info...)
> Let R(1/5 , -2) be a point on the parabola y^2 = 20x.
>
> from part (a) we find out the equation of the focal chord is:
> 5x-12y-25 = 0.
>
> (b) the co oridnates of Q where the focal chord cuts the directrix
> is (-5,
> -4 1/6)
>
> (c) the area of the triangle OFQ where O is the origin, F is the
> focus is
> 10 5/12 units squared.
>
> (d) te perpendicular distance from the chord to the point P (-1,
> -7) is 4
> 2/13 units.
>
> (e)! the bit i couldn't get...
> Hence find the area of triangle PQR. (answer 11.7 units^2

-------

Q8) find the equation of the parabola with axis parallel to the y
> axis and
> passing through points A(0, -2) B(1, 0) C ( 3, -8)


------

3a) find the equation of parabola with axis parallel to yaxis passing
> through a(0,3) b(1, 0 ) c (2, -1)

 

[Riviet]

(e) Area of triangle = 1/2 x (base) x (perpendicular height)

= 1/2 x (distance from P to Q) x (perpendicular distance from the chord to the point R)

You have the perpendicular distance found in (d) and just need to apply the distance formula for P and Q.

 

[airie]

Q8) find the equation of the parabola with axis parallel to the y
> axis and
> passing through points A(0, -2) B(1, 0) C ( 3, -8)


3a) find the equation of parabola with axis parallel to yaxis passing
> through a(0,3) b(1, 0 ) c (2, -1)

The most straightforward solution for both, sub all three points into the general formula (x-h)²=4a(y-k) and solve simultaneously for a, h, k

 

[-pari-]

ahhh....okay...great i'll try all three questions again and see how they work out

thanks riviet & airie

 

[jwarnott]

really sorry to be a pest and take over this thread but i didnt see the point in clogging up the forum with yet another locus question.

Find the expression for the distance PA is A is the point (-3,4) and P is any point (x,y).
show that the locus of the point P which moves so that the distance PM from the axis is equal to its distance PA from the point (-3,4) is 8y=x^2+6x+25

i am cluless how to go about this trying to work is out with the equation form the line. also baffled by this Q; find the focus and directrix of y=-1/2x^2

y=-1/2x^2
hence x^2=-2y

i am confused about this step and how is manages to become -2y... the rest is self explanitory.

if i could get an answer asap it would be much appreciated.

ps. sorry if i have not conducted myself in the proper manner. im new to the board and just getting my wits about me.

 

[SoulSearcher]

Multiply y = -x²/2 by -2, to get -2y = x²

 

[jwarnott]

wow, how embarrasing... haha, bombarded by so much work right now. dont judge me. still can't figure out my first question.

 

[jwarnott]

bump. sorry to be rude, getting anxious.

 

[-pari-]

Find the expression for the distance PA is A is the point (-3,4) and P is any point (x,y).
show that the locus of the point P which moves so that the distance PM from the axis is equal to its distance PA from the point (-3,4) is 8y=x^2+6x+25

which axis? x or y?

 

[Riviet]

Find the expression for the distance PA is A is the point (-3,4) and P is any point (x,y).
show that the locus of the point P which moves so that the distance PM from the axis is equal to its distance PA from the point (-3,4) is 8y=x^2+6x+25

which axis? x or y?

It's the x-axis (don't ask me why, I just know ).

So the distance PA = sqrt{(x+3)² + (y-4)²}. Now the distance from the x-axis to P is just y (we are making a big assumption that it's the perpendicular distance from the axis).

So distance PA = y

sqrt{(x+3)² + (y-4)²} = y

Squaring both sides and expanding,

x² + 6x + 9 + y² -8y + 16 = y²

.'. 8y = x² + 6x + 25