the parametrics animal (1 Viewer)

[.ben]

some questions:

11. P is a variable point on the parabola x^2=-4y . The tangent from P cuts the parabola x^2=4y at Q and R. Show that 3x^2=4y is the equation of the locus of the mid-point of the chord RQ.

26. How many normals pass through (0,ka) , a point on the axis of the parabola x^2=4ay, for k>2 ? For k=3, find where the normal meets and parabola again.

30. The chord of contact of the tangents to the parabola x^2=4ay from the point P(x,y) passes through the point Q(0, 2a) . Show that the locus of the midpoint of PQ is the X-axis.

thanks

 

[haboozin]

some questions:

11. P is a variable point on the parabola x^2=4ay . The tangent from P cuts the parabola x^2=4ay at Q and R.

i dont get it?

P is on that parabola..

its tangent cuts the same parabola 3 times?

but a tangent is only ment to cut its equation once, by definition.

 

[Raginsheep]

Im pretty sure a tangent can cut it's original curve more than once

but that question seems a bit wrong. Sure it isn't something like cuts the curve x^2=-4ay?

 

[.ben]

000ooooOOOps! sorry it's actually x^2=-4ay

 

[haboozin]

some questions:

11. P is a variable point on the parabola x^2=4ay . The tangent from P cuts the parabola x^2=-4ay at Q and R. Show that 3x^2=4y is the equation of the locus of the mid-point of the chord RQ.

what happened to a? in that locus?

i get 3x^2 = 4ay

maybe a = 1?

 

[.ben]

oops i totally screwed teh qun. here's teh real thing:

11. P is a variable point on the parabola x^2=-4y . The tangent from P cuts the parabola x^2=-4y at Q and R. Show that 3x^2=4y is the equation of the locus of the mid-point of the chord RQ.

sorry

 

[Trev]

[ps. .ben when you corrected yourself again you made another mistake that P is a variable point on x²=-4y when it is x²=4y]
Equ. tangent is y=px-p²
Intersection of tangent and x²=-4y gives points Q and R:
x²=-4(px-p²)
x²+4px-p²=0
(x+2p)²=8p²
x=-2p +/- 2√2p (x coordinates of Q and R)
-4y=(-2p +/- 2√2p)²
-4y=4p² +/-8√2p + 8p²
y=+/-2√2p - 3p² (y coordinates of Q and R)
Midpoint = (-2p,-3p²)
p=(-x/2)
y=-3(x²/4)
4y=-3x²
Screwed up somewhere because it's negative but can't be bothered finding where...
EDIT: I used x²=4y as the parabola which P is on and x²=-4y the one which it intersects, swap it around and you should get the required result. Just follow the same process though. (Your latest post is still wrong though lol)

 

[Trev]

26)
Normal of P on x²=4ay is x+py=2ap+ap³
at (0,ka)
pka=2ap+ap³
k=2+p²
p=+/-√(k-2)
when k>2 p is undefined (negative square root) therefore no normals blah...

when k=3, p=+/-1. Since point P is (2ap,ap²) (p=1) then normal meets parabola again at (-2ap,ap²).

 

[.ben]

[ps. .ben when you corrected yourself again you made another mistake that P is a variable point on x²=-4y when it is x²=4y]
Equ. tangent is y=px-p²
Intersection of tangent and x²=-4y gives points Q and R:
x²=-4(px-p²)
x²+4px-p²=0
(x+2p)²=8p²
x=-2p +/- 2√2p (x coordinates of Q and R)
-4y=(-2p +/- 2√2p)²
-4y=4p² +/-8√2p + 8p²
y=+/-2√2p - 3p² (y coordinates of Q and R)
Midpoint = (-2p,-3p²)
p=(-x/2)
y=-3(x²/4)
4y=-3x²
Screwed up somewhere because it's negative but can't be bothered finding where...
EDIT: I used x²=4y as the parabola which P is on and x²=-4y the one which it intersects, swap it around and you should get the required result. Just follow the same process though. (Your latest post is still wrong though lol)

sorry for confusing you but it is x²=-4y. which explains the negative answer.
thanks for the answers!

 

[Trev]

30)
Equation chord of contact is xx₁=4a(y+y₁) through point (0,2a) gives:
0=4a(y+2a), y=-2a.
Midpoint (a,b) of (x,y) and (0,2a) is a=x/2, b=(y+2a)/2
Since y=-2a then b (y-coordinate of midpoint) = (-2a+2a)/2 = 0.
∴ midpoint lies on y=0 which is the x-axis.

 

[.ben]

alright thanks trev! do parametrics get any harder than this in the hsc? (excluding ones with external/internal division)

 

[Trev]

Not unless you do 4u.

 

[.ben]

ok cool thanks!