More Locus Problems - NEED HELP (1 Viewer)

[blackops23]

Hi here's a question where I was successfully able to get the answer however my solution was way too long. As with most math questions, there always seems to be an alternative ninja method that cuts the working out in half. If possible, I'd like someone to think of a ninja solution, because I'm out of ideas.

here's the question:

Q. Two points, P and Q, move on the parabola x^2 = 4ay so that the x-coordinates of P and Q differ by a constant value, 2a. What is the locus of M, the mid-point of PQ?

Solution:

Its too long to write here, so I will just state what I did.
I let P=(2ap,ap^2) and Q=(2aq,aq^2)
But x-coordinates of P and Q differ by (2a)
So P=(2ap,ap^2) BUT Q=(2ap - 2a, aq^2)

So Q=(2a(p-1), aq^2)
which means, 2aq = 2a(p-1)
therefore, q=p-1
then I subbed that value of q in aq^2
so Q=(2a(p-1), a(p-1)^2)
and P=(2ap,ap^2)

Using these two values, I found the midpoint which was:

M=( a(2p-1), [a(2p^2 - 2p + 1)]/2 )
then x = a(2p-1)
rearranging to make p the subject
p= (x+a)/2a

then I subbed that into y=[a(2p^2 - 2p + 1)] /2, which was the main reason why the solution was so freaking long.

Is there any alternative method to find the LOCUS of M, using a shorter more elegant method?

Thanks guys.

 

[ultraman8]

For tonnes of maths resources visit my sig. or profile.

 

[Trebla]

Q. Two points, P and Q, move on the parabola x^2 = 4ay so that the x-coordinates of P and Q differ by a constant value, 2a. What is the locus of M, the mid-point of PQ?

Given P(2ap, ap²) and Q(2aq, aq²) the midpoint M is given by
M(a(p + q), a(p² + q²)/2)
The condition imposed is 2ap - 2aq = 2a (here I've assumed p > q)
=> p - q = 1
From parametric coordinates of M
x = a(p + q) => p + q = x/a
y = a(p² + q²)/2 => p² + q² = 2y/a
But
(p + q)² = p² + q² + 2pq
(p - q)² = p² + q² - 2pq
Upon addition:
(p + q)² + (p - q)² = 2(p² + q²)
=> x²/a² + 1 = 4y/a

 

[blackops23]

Given P(2ap, ap²) and Q(2aq, aq²) the midpoint M is given by
M(a(p + q), a(p² + q²)/2)
The condition imposed is 2ap - 2aq = 2a (here I've assumed p > q)
=> p - q = 1
From parametric coordinates of M
x = a(p + q) => p + q = x/a
y = a(p² + q²)/2 => p² + q² = 2y/a
But
(p + q)² = p² + q² + 2pq
(p - q)² = p² + q² - 2pq
Upon addition:
(p + q)² + (p - q)² = 2(p² + q²)
=> x²/a² + 1 = 4y/a

Thanks mate, that was brilliant!

But here's another problem which I'm having trouble with:

Q: At a point P on the parabola x^2 = 4ay, a normal PK is drawn. From the vertex O, a perpendicular OM is drawn to meet the normal at M. Show that locus of M as P moves on the parabola is given by: x^4 - 2(a)(x^2)(y) + (x^2)(y^2) - (a)(y^3) = 0.

What I did was simultaneously solve the equation of the normal at P: x + py = 2ap + ap^3 ------1
with OM, which was y=px ----2

Which I got: M= { [ap(p^2 + 2)]/(p^2 + 1) , [ap^2(p^2 + 2)]/(p^2 + 1) }

and then.... I got stuck =(


If anyone could give me a few hinters or a solution, it would be immensely appreciated.

Thanks guys.

 

[Trebla]

Thanks mate, that was brilliant!

But here's another problem which I'm having trouble with:

Q: At a point P on the parabola x^2 = 4ay, a normal PK is drawn. From the vertex O, a perpendicular OM is drawn to meet the normal at M. Show that locus of M as P moves on the parabola is given by: x^4 - 2(a)(x^2)(y) + (x^2)(y^2) - (a)(y^3) = 0.

What I did was simultaneously solve the equation of the normal at P: x + py = 2ap + ap^3 ------1
with OM, which was y=px ----2

Which I got: M= { [ap(p^2 + 2)]/(p^2 + 1) , [ap^2(p^2 + 2)]/(p^2 + 1) }

x = ap(p² + 2) / (p² + 1)
y = ap²(p² + 2) / (p² + 1)
Clearly y = px => p = y/x
=> x = a(y/x)(y²/x² + 2) / (y²/x² + 1)
=> x² = ay(2x² + y²) / (x² + y²)
=> x⁴ + x²y² = 2ax²y + ay³
=> x⁴ - 2ax²y + x²y² - ay³ = 0

 

[blackops23]

Thanks for the solution, but here's another problem I'm having trouble with ( I really stink at locus...)

Q. Tangents are drawn to a parabola x^2 = 4y from an external point A(x1, y1) touching the parabola at P and Q.
(i) Prove that the midpoint, M, of PQ is the point (x1, 0.5*(x1)^2 - y1)

The only way I can think of doing that is solving the chord of contact with x^2 = 4y, to form a quadratic then using the quadratic formula to get the coordinates of P and Q, and then finding the midpoint.

There has to be a quicker way, because that method would take to long, if anyone could come up with something it would be greatly appreciated.

Thanks guys =)

 

[Trebla]

When you solve the quadratic equation, there will be two roots corresponding to the x-values of P and Q. When you find the midpoint, you essentially add those x-value roots and then divide by 2. The addition of these roots corresponds to - b/a of the quadratic equation. This allows you to find M without explicitly finding the x-coordinates of P and Q.

 

[blackops23]

Thanks mate, but now I'm having a lot of trouble with the second part of the question...

(ii) If A moves along the straight line y = x - 1, find the equation of the locus of M.

I've never come across a question of this type so I'm unsure what to do. All I know is, A=(x1, y1) and M=(x1. -0.5(x1)^2 - y1)
So A and M lie on vertical line x = x1

That's all I know, but I don't really know how to formulate a solution, any help would be amazing,

Thanks guys.

 

[Trebla]

If A lies on y = x - 1 then y₁ = x₁ - 1, use this to determine the Cartesian equation of M with 'parameters' x₁ and y₁

 

[blackops23]

If A lies on y = x - 1 then y₁ = x₁ - 1, use this to determine the Cartesian equation of M with 'parameters' x₁ and y₁

Ok thank you for that one, really simplified the problem for me =)

However there is one more problem, it's the last question in the Jones&Couchmann exercise for Locus relating to the parabola

Q. PQ is a focal chord in the parabola x^2 = 4ay. If M is the mid-point of the focal chord PQ, and a line through M parallel to the axis of the parabola meets the normal at P in A, find the locus of A.

Now what I did:

I found the coordinates of M: { a(p+q) , [a(p^2 + q^2)]/2 }
Now AM is parallel to y-axis, so AM: x=a(p+q), so abcissa of A is a(p+q)
But A lies on the normal at P: x + py = 2ap + ap^3
So I substituted the x=a(p+q) in the equation of the normal:

a(p+q) + py = 2ap + ap^3
ap + aq + py = 2ap + ap^3
therefore
y= (ap + ap^3 - aq)/p

And that's where I got stuffed.

Any hinters or clues into how to find the LOCUS of A would be greatly appreciated.

Thanks guys.

 

[deterministic]

Remember PQ is a focal chord. Use that to get a relation between p and q.

 

[blackops23]

Do I apply q=-1/p on x=a(p+q)????, or just the ordinate value?

x=a(p+q)
= a(p - 1/p)

y=a(p + p^3 -q)
= a(p + p^3 + 1/p)

err....

x/a = p - 1/p
(x/a)^2 = p^2 - 2 + 1/p^2
y/a = p + p^3 + 1/p

yep, its pretty obvious, I got no idea what the heck I'm doing. A little more direction would be absolutely lovely =)

Thanks guys.

 

[deterministic]

x/a=(p-1/p)
(x/a)^2=p^2+1/p^2-2
y/a= (p + p^3 - q)/p=(p + p^3 + 1/p)/p= 1+p^2+1/p^2

now you have common p^2+1/p^2 in equations of x and y, so i shall leave the rest for you to complete.

 

[blackops23]

x/a=(p-1/p)
(x/a)^2=p^2+1/p^2-2
y/a= (p + p^3 - q)/p=(p + p^3 + 1/p)/p= 1+p^2+1/p^2

now you have common p^2+1/p^2 in equations of x and y, so i shall leave the rest for you to complete.

THANK YOU, SO MUCH, looks like I was only a couple of steps away =(

Also could you give me advice, on what situations I would have to use the q=-1/p to find the LOCUS of a point?