Fitpatrick Ex 24c) Q. 26 Really hard parametric equation (1 Viewer)

[Gangstar1]

Can anyone do this question:

How many normals pass through (0,ka), a point on the axis pf the parabola x^2 =4ay, for k>2? For K=3, find where the normal meets the parabola again.

TY!!!~

 

[muzeikchun852]

i havent done parametric for a long time (since year 11), so im not quite sure whether im right or not. hope it helps.

 

[hscishard]

for the k>2 part,
you should get y1 = (k-2)a
When k> 2, y1 is positive. The parabola is symmertrical, for every y value, there are 2 x values hence there will be 2 normals that intersect at that point

y1 = a
Find x1
Subsititute it into the normal equation
Find where it intersects again by using simulatenous equations

it's pretty hard to explain without typing up the whole solution

 

[blackops23]

FUAAARKK this is a pretty hard 3U question, well at least I hope so

 

[Drongoski]

FUAAARKK this is a pretty hard 3U question, well at least I hope so

If you have Cambridge 3U Year 11(1999 Edition) there is a worked example on Page 335 on a similar type of problem. I think for k > 2, there are 3 normals, including the y-axis.

 

[Trebla]

Can anyone do this question:

How many normals pass through (0,ka), a point on the axis pf the parabola x^2 =4ay, for k>2? For K=3, find where the normal meets the parabola again.

TY!!!~

Using the usual parametric coordinates with parameter p, equation of normal is
x + py = 2ap + ap³
To pass through (0, ka)
apk = 2ap + ap³
ap(k - p² - 2) = 0
p = 0, ±√(k - 2)
These correspond to three points possible points of contact on the parabola

 

[hscishard]

Wow. Fitzpatrick also forgot the nomal at the origin. Lol

 

[xxxzxc]

first find the equation of the normal, then substitute (0, ka), make t the subject and there should be 2 solutions for the sqrt(k-2).
for the next part, simply substitute k=3 then use your previous 2 solutions (from the first part) to find the 2 t-values. substitute this to the coordinates (2at, at^2) and you should get (2a, a) and (-2a, a).

 

[Gangstar1]

ahh answer is 2; (+- (plus minus) 6a ,9a) i don't think anyone got it. 0.0

 

[hscishard]

Drongoski and Trebla got part 1
I'll finish off my part 2

y1 = a
Then x1 = +/-2a (sub a back into x^2=4a(y)

Normal: y= +/- 1 (x+/-2a) + a
Solve simulatenously with x^2 = 4ay
You will get +/- 6a
Then sub this back in and you'll get y= 9a

 

[rawrence]

ahh answer is 2; (+- (plus minus) 6a ,9a) i don't think anyone got it. 0.0

You got the equation of a normal at
x+py=ap³ + 2ap
subbing in the point (0,ka)
k = p² +2
p² = k-2
Since k>2
p² > 2
therefore there's 2 solutions, so 2 normals

EDIT: Looking at trebla's solutions, I don't actually see why there's only 2, it seems as though there is 3...I'd probably test the third one, just in case

 

[Trebla]

x + py = 2ap + ap³
To pass through (0, ka)
apk = 2ap + ap³
In this line do NOT divide both sides by ap because it assumes p is non-zero when in fact p = 0 is possible.

 

[rawrence]

x + py = 2ap + ap³
To pass through (0, ka)
apk = 2ap + ap³
In this line do NOT divide both sides by ap because it assumes p is non-zero when in fact p = 0 is possible.

No I completely agree with not dividing both sides, it's just the solutions say there's only 2..

 

[hscishard]

Hmmm. I think Trebla > Fitzpatrick because Trebla is most correct. Answers can be wrong you know