Parametrics - Locus (1 Viewer)

nrlwinner · Jan 4, 2010

Hi. I'm not sure how to answer questions involving midpoints where multiple parameters are used.

Eg. Locus of the midpoint of P and Q on x^2 = 4ay where the x-coordinate differ by 2a

shaon0 · Jan 4, 2010

nrlwinner said:
Hi. I'm not sure how to answer questions involving midpoints where multiple parameters are used.

Eg. Locus of the midpoint of P and Q on x^2 = 4ay where the x-coordinate differ by 2a

P(2ap,ap^2) and Q(2ap-2a,aq^2) ie. 2aq=2ap-2a => q=p-1 => q=(x-a)/2a
Midpt: (2ap-a,a/2(p^2+q^2))

x=2ap-a
x+a=2ap
p=(x+a)/2a ...sub into y equ

y=a/2(p^2+q^2)
y=(a/2)((x+a)^2/4a^2+((x-a)/4a^2))
y=(1/8a)((x+a)^2+(x-a)^2)
y=(1/4a)(x^2+a^2)
x^2=4ay-a^2

nrlwinner · Jan 5, 2010

q=(x-a)/2a

Can you eplain how you got that?

blackratpoo · Jan 5, 2010

i h8 parametrics. dont get it a bit. anyone got tips?

shaon0 · Jan 5, 2010

nrlwinner said:
q=(x-a)/2a

Can you eplain how you got that?

p-1=[(x+a)/2a]-1
p-1=(x+a-2a)/2a
p-1=(x-a)/2a

ninetypercent · Jan 5, 2010

i h8 parametrics. dont get it a bit. anyone got tips?

- remember the definition of the parabola - it may come in handy
- learn to eliminate parameters to form a relationship between x and y
- keep bashing the algebra
- draw a diagram!
- remember formulas such as midpoint formula

nrlwinner · Jan 5, 2010

Thanks I got it now. Also, I've been stuck on this one. I know the co-ordinates are right

$(\frac{ap^2(p+1)}{p^2+1},\frac{ap^3(p+1)}{p^2+1})$

Can someone help me turn that into a locus equation. I know I have to let the x-coordinate equal x and then make the subject p, then sub it into y, but is this algebra possible?

untouchablecuz · Jan 5, 2010

nrlwinner said:
Thanks I got it now. Also, I've been stuck on this one. I know the co-ordinates are right

$(\frac{ap^2(p+1)}{p^2+1},\frac{ap^3(p+1)}{p^2+1})$

Can someone help me turn that into a locus equation. I know I have to let the x-coordinate equal x and then make the subject p, then sub it into y, but is this algebra possible?

$\\x=\frac{a(p+1)}{p^2+1}p^2 \ (*)\\y=\frac{a(p+1)}{p^2+1}p^3=\frac{a(p+1)}{p^2+1}p^2\times p=xp\Rightarrow p=\frac{y}{x} \ (**)\\\text {Sub }(**) \text{ in }(*),\\x=\frac{a(\frac{y}{x}+1)}{(\frac{y}{x})^2+1}\times(\frac{y}{x})^2\\x^2(x^2+y^2)=ay^2(x+y)$

nrlwinner · Jan 5, 2010

Thanks. Can someone help me finish this one off as well.

It's a question that the ends of the normals of a focal chord meet at R. Find the locus of R.

I've got the coordinates to be

$(a(p-q),a(p^2+q^2+1))$

Can someone check if that's right and how I can finish the question off.

jet · Jan 5, 2010

Ok, heres how you go from there. (I didn't check your working up to there. Assume it's correct)

The tiny bit of information that you might have missed is that they lie on a focal chord.

Substituting the focus into the equation of a chord gives you the condition that pq = -1.

Now, you complete the square.

We know that y = a(p^2 + q^2 + 1)
= a[(p^2 - pq + q^2) + pq + 1]
= a[(p - q)^2 + pq + 1]
= a(p - q)^2 since pq = -1

Now we know x/a = p - q

Therefore y = x^2/a
x^2 = ay

Which is a parabola with focal length a/4 and vertex at the origin

In locus and the parabola, there will always be a small piece of information which makes the question solvable.

nrlwinner · Jan 6, 2010

Thanks for the help so far. I'm nearly finished parametrics. Just a couple of last questions.

Here's a question I simply cannot do.
Tangets are drawn to a parbola x^2=4y from an external point A(x1,y1) touching the parabola at P and Q. Prove that the midpoint, M, of PQ is the point

$(x1,\frac{1}{2}(x1^2-y1)$

Also. I cannot turn this point into a locus. There doesn't seem to be any other information in the question that can help me. My points are

$(\frac{ap(p^2+p-1)}{p^2+1},\frac{a(p^4+p^4+1)}{p^2+1})$

untouchablecuz · Jan 6, 2010

nrlwinner said:
Thanks for the help so far. I'm nearly finished parametrics. Just a couple of last questions.

Here's a question I simply cannot do.
Tangets are drawn to a parbola x^2=4y from an external point A(x1,y1) touching the parabola at P and Q. Prove that the midpoint, M, of PQ is the point

$(x1,\frac{1}{2}(x1^2-y1)$

Also. I cannot turn this point into a locus. There doesn't seem to be any other information in the question that can help me. My points are

$(\frac{ap(p^2+p-1)}{p^2+1},\frac{a(p^4+p^4+1)}{p^2+1})$

$\\\text {Chord of contact through points P and Q is } xx_1=2(y+y_1) \text { (this can be memorised)}\\\text {Solving simultaneously with the equation of the parabola,}\\2x^2-4x_1x+8y_1=0\\\text{The roots of this equation, }\alpha \text{ and }\beta \text{, are the } x\text {-coordinates of P and Q}\\\therefore \text {The }x \text{ coordinate of the midpoint is }M_x=\frac{\sum \alpha}{2}=\frac{-\frac{-4x_1}{2}}{2}=x_1\\\text {Sub } M_x \text{ into the equation for the chord of contact to find the }y\text {-coordinate,}\\x_1x_1=2(M_y+y_1)\Rightarrow M_y=\frac{1}{2}(x_1^2-2y_1)$

did you mistype the y coordinate for the first Q?

also, can you check over the coordinates for the second Q, i have a feeling you mistyped them

nrlwinner · Jan 6, 2010

I've already got the first question. Thanks.

The y coordinate for the 2nd question is meant to be

$(\frac{a(p^4+p^3+1)}{p^2+1})$
If that doesn't help, the question is.
If PN is a normal to the parabola x^2=4ay at a variable point P and SN is drawn through the focus S parallel to the tangent at P to cut the normal at N. Find the locus of N

jet · Jan 6, 2010

You made a mistake with the coordinates, I'm not sure where. This is the correct answer:

$\text{The line through the focus is parallel to the tangent, therefore it has gradient } p \\ \therefore y - a = p \left (x - 0 \right ) \\ y = px + a \\ \text{The normal has equation } x + py = ap^3 + 2ap \\ \text{Solving the two together, } x + p^2x + ap = ap^3 + 2ap \\ \left(1 + p^2 \right )x = ap\left (p^2 + 1 \right ) \\ x = ap \\ \therefore y = a\left (p^2 + 1 \right ) \\ \text{Solving for } p, p = \frac{x}{a} \\ \therefore y = a\left (\frac{x^2}{a^2} + 1 \right ) \\ ay = x^2 + a^2 \\ x^2 = a\left (y - a \right ) \\ \text{Which is a parabola, with vertex on the original focus and focal length } \frac{a}{4}$

Parametrics - Locus (1 Viewer)

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