Parametric Equations of the Parabola (1 Viewer)

[FDownes]

I'm having trouble with a particular parametric parabola question. Can someone go through them with me so I can understand? Here it is;

Prove that SP = a(p² + 1) units [where S = (0, a) or the focus, and P = (2ap, ap²)] and hence that SZ² = OS x SP [where O is the origin].

 

[tommykins]

SP
= sqrt [ (2ap)² + (a-ap²)² ]
= sqrt [ 4a²p² + a² -2a²p² + a²p^4 ]
= sqrt [ 2a²p² + a² + a²p^4 ]
= sqrt [ a²(2p²+1+p^4) ] -> brackets factorises to (p²+1)²
= sqrt [ a²(p²+1)² ]
= a(p²+1)

Second question is kinda confusing, what's Z?

 

[FDownes]

For Z (I'm assuming you mean the first question);

= (0) - px - ap²
= px = ap²
= x = ap

therefore Z = (ap, 0)

 

[tommykins]

Drawing up the diagram.

OS = a
SP = a(p²+1) -> from above.

OS.SP = a* a(p²+1) = a²(p²+1)

SZ = sqrt [ (ap)² + a² ]

But since it's SZ²

it becomes SZ² = a²p² + a² = a²(p²+1) = OS.SP

 

[FDownes]

I'm having trouble with part iv and v of the same question. Can I see your answers?

 

[tommykins]

You didn't give me part IV and V

 

[FDownes]

Sorry, got halfway through typing it up and accidentally deleted it. I'll type up the full question...

The equation of the tangent at P(2ap, ap²) on the parabola x² = 4ay (whose vertex is 0) is y - px + ap² = 0. The tangent meets the x-axis at Z (ap, 0) and the y-axis at T (0, -ap²). PN is drawn perpendicular to the axis of the parabola meeting it at N (0, ap²), and PM meets the directrix at right angles in M.

iv) Prove that the tangent PT makes equal angles wih the axis of the parabola and wih PS (HINT: show triangle PST is isosceles)

v) Prove the lines SM, TP intersect at right angles at Z.

 

[tommykins]

Stuffed up post. =\

 

[tommykins]

iv ) Draw a diagram of it, and I will take you step by step through the proofs.

PST form a triangle, and from the diagram you can see that ST should equal SP.

Distance formula ST and you will get a(p²+1) = SP -> from above.

Now, let < NPT = < SPT = @ (theta)

 

[tommykins]

Oh my god, the posting thing stuffed up. Finlay add me on MSN, i'll pm you it.

 

[FDownes]

Thanks for all your help Tommy.

Now, time for another question;

P (2p, p²) is a variable point on the parabola x² = 4y, whose focus is S and vertex O. M and N are the midpoints of SP and OM respectively.

a) Show the locus of M is the parabola y = 1/2(x² + 1) and find the locus of N.

I'm pretty sure that x should equal p, but beyond that I'm lost. Can anyone help?

 

[tommykins]

P = (2p,p²)
S = (0,1)

x = x1 + x2 / 2 = 2p+0 /2 = p
y = y1 + y2 / 2 = p² + 1 / 2 = x²+1 / 2 = 1/2(x²+1)

 

[FDownes]

How do you establish that the locus is (0, 1)? Is it just an assumption?

 

[tommykins]

S = focus.

x² = 4ay, since the parabola eqn given is x² = 4y, a = 1.

Focus has coordinates (0,a) which then becomes (0,1) and you just work with that.

 

[FDownes]

Ooooooooh... I get it now. I have no idea why I didn't see that earlier.

Thanks very much.

 

[tommykins]

N is midpoint of OM.

M = (p, (p²+1)/2)
O = 0

Thus N has co-ordinates (p/2, (p²+1)/4)

x = p/2
y = ((p²+1)/4)

obviously you can't sub anything in y since x = p/2, so i'm guessing you just sqaure x to make it x² = p²/4

thus p² = 4x²

y = (4x²+1)/4
4y = 4x² +1

since i've given you both locus's, you can try to get the focus/whatever yourself.

 

[FDownes]

I'm having problems with another question, largely because I'm having a bit of trouble visualising what the question asks. Here's the question;

The equation of the tangent at a variable point T (2t, t²) on x² = 4y is y - tx + t² = 0. This tangent meets the tangent at the vertex O in R and the axis in Q.

a) Find the locus of M, the midpoint of QR.

b) If ORPQ is a rectangle, find the locus of P. What relation is there between the focal lengths of these loci?


Should the tangent at O simply be considered to be y = 0? And in that case, should I simply substitute y = 0 in to the tangent's equation to find R? By axis, does the question refer to the x- or y-axis, or the axis of the parabola itself (i.e. x = 0)? Here's my working;

For R;
= (0) - tx + t² = 0
= tx = t²
= x = t
= R = (t, 0)

For Q;
= y - t(0) + t² = 0
= y = - t²
= Q = (0, -t²)

 

[FDownes]

Tommy saved me again with the last question, but now I have another;

The tangent at T (4t, 2t²) on the parabola x² = 8y has equation y - tx + 2t² = 0.

a) This tangent meets the parabola x² + 4y = 0 in P and Q. Show that the abscissae x₁ and x₂ of the points P and Q are the roots of the quadratic equation x² + 4tx - 8t² = 0 and write down the value of 1/2(x₁ + x₂)

b) If M (x, y) is the midpoint of PQ, prove that x = -2t, and find y in terms of t. Hence find the equation of the locus of M, as T varies.

 

[FDownes]

Ignore the last question, I've got a new one. Again, I'm having problems visualising exactly what the question asks. Here it is;

PQ, a variable chord of the parabola x² = 4y, subtends a right angle at the vertex O.

a) If p and q are parameters corresponding to P and Q, show that pq = -4.

b) If the tangents at P and Q intersect at T, find the locus of T.

c) If M is the midpoint of chord PQ, determine the locus of M.


(It'd be helpful if the worksheet that I'm getting these questions from actually gave the answers; I have no way of checking whether I'm right or not.)

 

[independantz]

a) let the gradient of OP = -1/OQ or vice versa.

Mop=ap^2-0/2ap-0=p/2
similarly,Moq=q/2
therfore,p/2=-2/q => pq=-4

b)eqn of tangent at p=> y=px-ap^2 ( pretty straight forward to derive)

similarly, tangent at q => y=qx-aq^2
qx-aq^2=px-ap^2
(q-p)x=a(q^2-p^2)
=a(q-p)(p+q)
therefore, x=a(p+q), p doesnt equal q
sub back into tngt eq: y=p(a(p+q))-ap^2
y=apq, but pq=-4
y=-4a
=-4 ,since a =1
Therefore the locus of p is the line y=-4 since x is variable and y is constant.

c)M(x+x/2, y+y/2)

2p+2q/2,q^2+p^2/2

Therefore, x=p+q and y=((p+q)^2-2pq)/2=((p+q)^2+8)/2

i.e.y=(x^2+8)/2=x^2/2+4