what the hell (1 Viewer)

[fashionista]

dayem this question!!!!
grrrr
ok see if u can get the answer the book says (as opposed to mine)

P is a variable point on the parabola x=2t, y=tt with focus S and vertex A. Q is the mid-point of SP and r the midpoint of AQ. Find the Cartesian equation odf the locus of R. Show that the locus is a parabola and find it's vertex and focus.

grrrrrrrrrrrrr the book says the locus of R is xx=2(y-1/2) but I say it's xx=y-1/4

 

[CM_Tutor]

I agree with you, I get y = x^2 + 1 / 4

 

[dawso]

yeah ilma,
i got the same as u, book must be pretty lame, is that from fitz.

 

[fashionista]

yuppers it is indeedie

 

[fashionista]

okay this one i just cant do
P is a variable point on the parabola xx=-4y. the tangent from P cuts the parabola xx=4y at Q and R. Show that the equation of locus of the midpoint of the chord RQ is 3xx=4y.

 

[dawso]

i dunno, bout time ya did ya own questions and stopped askin em here, lol
na, ill try it l8r, seems pretty easy, i can see the graph in ma head, might cum out, mite not

 

[fashionista]

shudup dawso.
GAQAAARBJHDJKGSHFGLK!!!!!!!!!!!!!!!!!!
I hate parametric questions!!!!!! (that i cant get)
like this one:
Through the 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 at a fixed point K.

and this one

If two tangents to the parabola x=2at, y=att meet at right angles at Q and the normals at their points of contact meet at R, prove that QR is parallel to the axis of the parabola.

Can someone please explain these to me?????
Ur help is muchly appreciated

 

[AGB]

for the first question... (this is a theorem, where if pq = -1, then the chord PQ is a focal chord)

the gradient of the chord PQ is: 1/2 (p + q) .... (using gradient formula)

equation of the chord PQ is: y - .5(p + q)x + apq = 0 .... (using the equation y - y1 = m(x - x1) and subbing in y1 = ap^2 and x1 = 2ap)

therefore if we sub in the point F (0,a) into the above equation....

a - .5(p + q)*0 + apq = 0
apq = -a
:. pq = -1

and as pq = -1 (given in question), then the chord PQ always passes through the point F (0,a), which is on the axis of the parabola (x = 0)

 

[AGB]

i am not sure about the second question....

one x coordinate is a(p + q) and the other x coordinate is -apq(p + q), and if QR is parallel to the y axis, then dont the x coordinates have to be the same?? therefore, QR is not parallel to the y axis??

 

[dawso]

dont worry ilma, i hat parametric shit too, we can just get o for that question together

 

[AGB]

can someone like cm_tutor or one of the other smarties here please try the second question?? or at least say if it is a valid question?? thanks

 

[fashionista]

yeah where is he???

 

[CM_Tutor]

Originally posted by fashionista
If two tangents to the parabola x=2at, y=att meet at right angles at Q and the normals at their points of contact meet at R, prove that QR is parallel to the axis of the parabola.

Let's take the two points on the parabola be M(2am, am²) and N(2an, an²)

The tangents at M an N are, respectively, y = mx - am² and y = nx - an². They meet when mx - am² = nx - an²
ie x(m - n) = a(m² - n²) = a(m + n)(m - n)
ie x = a(m + n) - which is the x coordinate of Q

Note that since we are trying to prove that QR is vertical, we only need to show that Q and R have the same x coordinate.

Now, the normals at M is x + my = 2am + am³ ______ (1)
and similarly, the normal at N is x + ny = 2an + an³ ______ (2)
We seek the x value, so we look at (1) * n - (2) * m:
nx + mny - (mx + mny) = 2amn + am³n - (2amn + amn³)
x(n - m) = amn(m² - n²) = amn(m + n)(m - n)
x = -amn(m + n) is the x coordinate of R

Clearly, AGB is right. I should have done it on paper first - Oops. The question makes no sense as it stands. What it should say is that MN is a focal chord. Then, mn = -1, and the x coordinate of R becomes a(m + n), which is the same as the x coordinate of Q, and so QR is vertical (and hence parallel to the axis of the parabola).

 

[CM_Tutor]

Originally posted by AGB
for the first question... (this is a theorem, where if pq = -1, then the chord PQ is a focal chord)

the gradient of the chord PQ is: 1/2 (p + q) .... (using gradient formula)

equation of the chord PQ is: y - .5(p + q)x + apq = 0 .... (using the equation y - y1 = m(x - x1) and subbing in y1 = ap^2 and x1 = 2ap)

therefore if we sub in the point F (0,a) into the above equation....

a - .5(p + q)*0 + apq = 0
apq = -a
:. pq = -1

and as pq = -1 (given in question), then the chord PQ always passes through the point F (0,a), which is on the axis of the parabola (x = 0)

Sorry, AGB but this is wrong. You have assumed that the chord is a focal chord. This is not what the question said, and if it had the question would be completely pointless - obviously all focal chords pass through the same point on the axis - namely the focus, because that's what a focal chord is!. The question actually said that:

Originally posted by fashionista
Through the 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 at a fixed point K.

So, let P be (2ap, ap²), and Q be (2aq, aq²).

As AGB said, the gradient of the chord PQ is (p + q) / 2, and the chord PQ is y = (p + q)x / 2 - apq.

Now, we also know that AP and AQ are perpendicular. Since A is just the origin, we can easily show that the gradients of AP and AQ are p / 2 and q / 2, respectively.
m(AP) * m(AQ) = -1
(p / 2) * (q / 2) = -1
pq = -4.

Put this into the equation of PQ when x = 0, and we get that the y intercept is:
y = (p + q) * 0 / 2 - a(-4) = 4a.

So, the y-intercept of PQ (ie K) is (0, 4a), for all p and q, which is a fixed point as required.

 

[CM_Tutor]

Originally posted by fashionista
P is a variable point on the parabola xx=-4y. the tangent from P cuts the parabola xx=4y at Q and R. Show that the equation of locus of the midpoint of the chord RQ is 3xx=4y.

Let P, the point on x²= -4y, be (2p, -p²). It follows that the gradient of the tangent at P is -p, and its equation is
y = -px + p² ____ (1).
Solving this sultaneously with the second parabola y = x² / 4 _____ (2)
using (1) = (2), we get x² = -4px + 4p²
ie x² + 4px - 4p² = 0, the solution of which is the x coordinates of Q and R. We don't actually care what these are, we only want there midpoint, which is the sum of the roots divided by 2, ie -b / 2a
So, X coordinate of R is -4p / 2(1) = -2p

Rearranging (1) into x = (p² - y) / p and substuting this into (2), we get:
4y = (p² - y)² / p²
4p²y = p⁴ - 2p²y + y²
y² - 6p²y + p⁴ = 0
Again, we don't care about the solutions, only their average.
Thus, y coordinate of R is -(-6p²) / 2(1) = 3p².

Thus, R is at (-2p, 3p²)

The locus of R is then x = -2p ______(A)
y = 3p² ______(B)
Making p the subject of (A) and substituting into (B), we get the locus of R is:
y = 3(-x / 2)² = 3x² / 4
ie 4y = 3x² as required.