Maths Questions- Parametrics (1 Viewer)

[greenlemings]

Hi, I need some help on parametrics, not very good at it yet ^^
1.i) PQ is a focal chord to the parabola x^2= 4ay, find the locus of the mipoint M of PQ
1.ii) Hence, find locus of intersection K of tangents P and K, and show that MK is always parallel to the axis.

2. Find and describe the locus of the midpoint M of the chords cut off a parabola x^2= 4ay by the lines parallel to y= x.

If you could comment the answers and the solutions that would be very helpful.
Thanks everyone in advance for your help

 

[gcalway]

Hi greenlemings,

I confess to being a newby both to this forum and parametric equations, however my study to date leads me to offer the following solutions and also to refer you to this document which I have found invaluable in my learning www.nointrigue.com/docs/notes/maths/maths_parametrics.pdf

Question 1 (i)

For x^2 = 4ay the parametric equivalent is x = at , y = at^2

Given P(2ap, ap^2) and Q(2aq,aq^2) we can use the following to establish the mid-pint M

Mx = Qx + (Px – Qx)/2 and My = Qy + (Py – Qy)/2

Mx = 2aq + (2ap – 2aq)/2 which simplifies to Mx = a(p+q)

My = aq^2 + (ap^2 – aq^2)/2 which simplifies to My = (ap^2 + aq^2)/2

The parametric equations for the locus of M are therefore x = a(p+q) and y = (ap^2 + aq^2)/2

The tangents associated with the focal chord intersect at right angles and therefore pq = -1 (Refer PDF above).

x^2 = a^2(p+q)^2 Squaring the x equation

x^2 = a^2(p^2 + 2pq + q^2)

x^2 = a(ap^2 + aq^2 + 2apq) substituting for y = (ap^2 + aq^2)/2 & pq = -1

x^2 = a(2y – 2a) => y = x^2/2a + a

Question 1 (ii)

The intersection of the tangents occurs at K(a(p+q), apq) - (Refer PDF above).

The gradient of the line MK is given by”

m = (My – Ky)/(Mx – Kx)

Without completing the substitutions it is evident that the denominator (a(p+q – a(p+q)) = 0

This indicates an undefined value for m which indicates that MK is vertical and therefore parallel to the y-axis.

Question 2

Applying similar techniques to those used in Question 1 we can find that the mid-point of the points P(2ap, ap2) and Q(2aq,aq2) is:

M(a(p+q), (ap^2 + aq^2)/2)

We are asked to consider all lines parallel to y = x which implies all lines having gradient 1.

Using the gradient formula (Rise/Run) gives:

(aq^2 – ap^2)/(2aq – 2ap) = 1 which simplifies to (p + q) = 2

From the co-ordinates of M and substituting for (p+q) we can say that x = a(p+q) => x = 2a which is a vertical line.


I hope that helps,

Graeme

 

[greenlemings]

thanks so much!!!