Parametrics qu (1 Viewer)

[VenomP]

P is a variable point (2t, t^2) on the parabola x^2=4y, whose vertex is O.
M, N, are the feet of the perpendiculars from P on to the axis and directrix respectively.

R is the midpoint of OP and L is the midpoint of MN. Find the locus of:

i) R

ii) L

...why must parametrics be so hard... :'(

 

[alakazimmy]

P is a variable point (2t, t^2) on the parabola x^2=4y, whose vertex is O.
M, N, are the feet of the perpendiculars from P on to the axis and directrix respectively.

R is the midpoint of OP and L is the midpoint of MN. Find the locus of:

i) R

ii) L

...why must parametrics be so hard... :'(

M and N are the feet of the perpendiculars from to the axis and directrix.
Thus, co-ordinates of them are:
M(2t,0) and N(2t,-1)


i) R is the midpoint of OP.
So we can use the midpoint formula on the points O(0,0) and P(2t,t²)
So, the midpoint R is (t,t²/2)

The locus of R in parametric form becomes:
x = t, and y = t²/2
To convert to Cartesian, remove the parameter t.
x²= t²and 2y = t²Therefore, the locus of R is the parabola x²= 2y, which goes through the origin and has focal length of ¹/₂


ii)L is the midpoint of MN
Again, we use the midpoint formula, but on the points M(2t,0) and N(2t,-1)
So the midpoint L is (2t, -¹/₂)

The locus of L in parametric form becomes:
x = 2t and y = -¹/₂There's no restriction on t, so all values of x are in the domain of the locus. However, y is restricted.
So the locus of L is the straight line y = -¹/2

 

[jet]

What do you mean by the axis? That could mean the axis of the parabola.

 

[VenomP]

Yeah, the 'axis' is a bit ambiguous, but that's all the question says.

Okay, thanks a lot for that alakazimmy. That's really helped.

I've moved onto some of the tougher questions now. And I have another one i need help with. It involves solving simultaneous equations mostly, but its proving quite difficult.

11. Write down the equation of the chord joining the points P (2ap, ap^2) and Q (2aq, q^2) on the parabola x^2 = 4ay, and show that if PQ is a focal chord, then pq = 1. no problems here

i) Find the equation of the tangent at P and the co-ordinates of T, the point of intersection of the tangents at P, Q. Hence determine the equation of the locus of T as P, Q vary.

ii) Find the equation of the normal at P and the co-ordinates of N, the point of intersection of the normals at P, Q. Hence determine the equation of the locus of N.

Answers are:

11. (i) T is {a(p + q), apq}; locus of T is y = -a (the directrix)

(ii) N is {-apq(p + q), a(p^2 + pq + q^2 + 2)}; locus of N is x^2 = a(y - 3a)

 

[alakazimmy]

Woo, more parametrics.. -_-

So, the equation of the chord PQ should be:

y = ^x(p+q)/₂ - apq.
If PQ is focal chord, pq = -1 (I think you typed that part up wrong)

i) Find the equation of the tangent at P and the co-ordinates of T, the point of intersection of the tangents at P, Q. Hence determine the equation of the locus of T as P, Q vary

First, we need the equations of tangents.
^dy/_dx = ^2ap/_2a= p
Hence, equation of tangent is: (y - ap²) = p(x - 2ap)
y = px - ap²
Similarly, for Q, tangent is y = qx - aq²

So, to find T, we interest these two lines.
That is, find x for : px - ap² = qx - aq²px - qx = ap²- aq²
x(p - q) = a(p - q)(p + q)
Hence, the x-coordinate of T = a(p+q)

Subbing this value of x into either tangent for P or Q, you'll get the y value of T.
y = p[a(p+q)] - ap²
= ap² + apq - ap²
= apq

Thus, T is {a(p + q), apq}

From before, we know that pq = -1.
So T becomes {a(p + q), -a}, with no restriction on the x-value. The y-value is restricted to -a, so equation of the locus is: y= -a, or the directrix.


ii) Find the equation of the normal at P and the co-ordinates of N, the point of intersection of the normals at P, Q. Hence determine the equation of the locus of N.

We have from part i) that ^dy/_dx = p, so the gradient of the normal would be -¹/_p

Equation of the normal at P is: (y - ap²) = -¹/_p(x - 2ap)
py - ap³ = -x - 2ap
x + py = 2ap + ap<sup>3

</sup>Likewise, for Q, the normal is: x + qy = 2aq + aq3

Solving these 2 simultaneously to find y-coordinate of N:
- py + 2ap + ap³ = - qy + 2aq + aq³
y(p - q) = 2ap + ap³ - 2aq - aq³
y(p - q) = 2a(p - q) + a(p - q)(p²+pq+q²)
Hence, y = a(2 + p²+pq+q²)

Substitute back into either normal equation to get x.
x + pa(2 + p²+pq+q²) = 2ap + ap³
x = 2ap + ap³ - 2ap - ap³ - ap²q - apq²
x = -apq(p + q)

Hence, N is {-apq(p + q), a(2 + p²+pq+q²)}

Finally, we can get the locus. Remembering that pq = -1, we can change the x and y-values.
x = a(p + q)
y = a(1 + p² + q²)

Here, it's quite tricky to eliminate the variables p and q.
x² = a²(p + q)²
y = a(1 + (p + q)² - 2pq) = a(3 + (p + q)²)

Note that ^{x^2}/_a^2 = (p + q)².

Hence, the locus is: y = a(3 + ^{x^2}/_a^2)
y = 3a + ^{x^2}/_a
y - 3a = ^{x^2}/_a
Hence, x² = a(y - 3a)


Phew... what a longgggg question.

 

[VenomP]

Thanks alakazimmy, that's brilliant. I'm finally starting to understand these questions - slowly but surely. I do have a couple more, but I think I might work on them a bit longer before posting them here.

Thanks for all your help so far, much appreciated.