calculus - locus help needed (1 Viewer)

[Petinga]

1. Find the co-ordiantes of 3 points on the parabola x^2=4y such that the normals through these points pass through the point (-12,15).

2. Throught he vertex A of a parabola, chords AP and AQ are drawn at right angles to one another. Show that for all positions of P, PQ cuts the axis of the parabola in a fixed point K.

3.P (2at,at^2) is any poiunt on the parabola x^2=4ay. The normal to the parabolaat P meets the y-axis at G and the x - axis at H. Q(u,v) is the forth vertex of the rectangle GOHQ. show that u=vt

4. The normal at any point P (2at,at^2) on the parabola x^2=4ay cuts the y-axis at Q and is produced to R so that PQ=QR.
a) Express the coordinates of R in terms of t.
b)Find the cartesian equation of the locus of R.

 

[who_loves_maths]

hi Petinga,

Part (1):

the parabola x^2 = 4y has parameters: T(2t, t^2) where T is any point on the parabola, and where focal length a = 1.

gradient at T = t ;
ie. equation of Normal is given by: y - t^2 = (2t - x)/t

---> ty - t^3 = 2t - x ---> Equation: ty + x - 2t - t^3 = 0

the point P(-12, 15) satisfies the equation of the normal:

15t - 12 - 2t - t^3 = 0 = t^3 - 13t + 12

by inspection, t = 1 appears immediately as a roots of this equation since (1 - 13 + 12) = 0

hence, factorise the equation:

t^3 - 13t + 12 = (t -1)(t^2 + t -1) = (t -1)(t + (1 -sqrt5)/2)(t + (1 +sqrt5)/2)

ie. the three points, of the form {2t, t^2}, are as follows:

{2, 1}, {(sqrt5 - 1)/2, (3 - sqrt5)/2}, and, {(-sqrt5 - 1)/2, (3 + sqrt5)/2}


Part (2):

since you have not specified the equation or the position nor type of the parabola in question, i will have to assume it is a parabola in the standard position with equation: x^2 = 4ay
in which case, the vertex of the parabola is: (0, 0)

defined the points P and Q parametrically as: P{2ap, ap^2}, Q{2aq, aq^2}

ie. gradient of AP = (ap^2)/2ap = p/2 ; similarly, gradient of AQ = q/2

AP and AQ are perpendicular to each other:
(q/2)(p/2) = -1 -----> pq = -4

PQ has the equation: y - ap^2 = [(ap^2 - aq^2)/(2ap - 2aq)](x - 2ap)
---> y - ap^2 = (p + q)(x - 2ap)/2
---> 2y - 2ap^2 = (p + q)x - 2ap^2 - 2apq
---> (p + q)x - 2y - 2apq = 0

now, axis of parabola is given by: x = 0 ; so, y-intercept of QP is given by:

0 - 2y - 2apq = 0
---> y = -apq

but pq = -4 ; ie. y intercept is at y = 4a , with 'a' a constant.

hence, PQ cuts the axis of the parabola at a fixed point.


Part (3):

Normal of parabola is:
y - at^2 = (2at - x)/t -----> py - at^3 = 2at - x -----> ty + x - 2at - at^3 = 0

y-coordinate of G is given when you let x = 0:
ty = 2at + at^3

x-coordinate of H is given when you let y = 0:
x = 2at + at^3

ie. ty = x

in order to make the rectangle, Q must have the y-coordinate of G and x-coordinate of H:
ie. Q(u, v) ---> Q(x, ty) = Q(x, x/t) since ty = x

hence, v = u/t ---> u = vt


Part (4):

a)
Using derivations found in Part(3), we know the equation of the normal at any point and that it cuts the y-axis, when x = 0, at: y = 2a + at^2

now, if PQ = QR, then its equivalent to saying that "R cuts PQ externally in the ratio 2:1"

ie. using external division formula, we find the coordinates of R as:
R{(2(0) - (2at))/(2 -1), (2(2a + at^2) - (at^2))/(2 -1)}
= R{-2at, (4a + at^2)}

hence, coordinates of R, in terms of 't', is: R{-2at, (4a + at^2)}

b)
let: x = -2at -----> t = -x/2a
y = 4a + at^2

substitute t = -x/2a into expression for y:
y = 4a + a(x^2/4a^2) = 4a + x^2/4a

ie. cartesian equation of R is: 4ay = x^2 + 16a^2



hope this helps Petinga

 

[davidgoes4wce]

hi Petinga,

Part (1):

the parabola x^2 = 4y has parameters: T(2t, t^2) where T is any point on the parabola, and where focal length a = 1.

gradient at T = t ;
ie. equation of Normal is given by: y - t^2 = (2t - x)/t

---> ty - t^3 = 2t - x ---> Equation: ty + x - 2t - t^3 = 0

the point P(-12, 15) satisfies the equation of the normal:

15t - 12 - 2t - t^3 = 0 = t^3 - 13t + 12

by inspection, t = 1 appears immediately as a roots of this equation since (1 - 13 + 12) = 0

hence, factorise the equation:

t^3 - 13t + 12 = (t -1)(t^2 + t -1) = (t -1)(t + (1 -sqrt5)/2)(t + (1 +sqrt5)/2)

ie. the three points, of the form {2t, t^2}, are as follows:

{2, 1}, {(sqrt5 - 1)/2, (3 - sqrt5)/2}, and, {(-sqrt5 - 1)/2, (3 + sqrt5)/2}


hope this helps Petinga

You have done the factorising in this question wrong

t^-3-13t+12=(t-1)(t^2+t-12)

which is further broke down to (t-1)(t+4)(t-3)

which gives t=1,-4,3
With a=1, using the (2at,at^2)
t=1 (2,1)
t=-4 (-8,16)
t=3 (6,9)