Parametrics Questions [Locus and the Parabola] (1 Viewer)

[king_of_boredom]

hey guys, just need help on a few questions:

A) find the equation of the locus of point R that is the intersection of the normals at P(2p,p^2) and Q(2q,q^2) on the parabola x^2=4y, given that pq=-4.

B) Given that P(2ap, ap^2) and Q(2aq, aq^2) lie on the parabola x^2=4ay, chord PQ subtends a right angle at the origin;

i) show pq=-4,

ii) find the equation of the locus of the midpoint of PQ,

iii) show that this locus is a parabola, and find its vertex and focal length.

thx in advance.

 

[who_loves_maths]

hi king_of_boredom,

Question A):

the equations of normals are:

1) p^2 - (1/p)(x -2p) = y ; and, 2) q^2 -(1/q)(x -2q) = y {you can derive these easily yourself}

equating them: p^2 - (1/p)(x -2p) = q^2 -(1/q)(x -2q) ---> (p-q)x = -pq(p^2 -q^2) ; where -pq = 4 ;

---> (p-q)x = 4(p-q)(p+q) ---> (p does NOT =q) x = 4(p+q)

now, substitute this back into either one of the equations for the normal to obtain a value for the y-coordinate:

y =p^2 - (1/p)((4p + 4q) -2p) = p^2 -2 -4q/p ; then substitute -4 = pq in:

y =p^2 -2 + (pq)q/p =p^2 + q^2 -2 ; but (p^2 +q^2) = (p+q)^2 -2pq =(p+q)^2 +8;

hence, y = (p+q)^2 +8 -2 = (p+q)^2 +6 ; but we already found that x= 4(p+q):

so, substituting 'x' one last time, y = (x^2)/16 + 6 is the locus of the point of intersection of the normals.


Question B):

Part (i)

if the chord PQ subtends a right angle at (0, 0), then it's the same as saying that the line OP is perpendicular to the line OQ; in which case you only need to worry about the gradients of these two lines:

gradient of line OP = (ap^2 -0)/(2ap -0) = (p^2)/(2p) = (1/2)p
similarly, gradient of line OQ = (1/2)q

since both lines are perpendicular, then their gradients multiply to give -1.

ie. ((1/2)p)((1/2)q)= -1 ---> pq = -4


Part (ii)

x-coordinate of the midpoint is given by: (2ap + 2aq)/2 = x = a(p + q)
y-coordinate of the midpoint is given by: (ap^2 + aq^2)/2 = y = (a/2)(p^2 + q^2)

remember that (p^2 + q^2) = (p+q)^2 - 2pq; where pq = -4;
ie. (p^2 + q^2) = (p+q)^2 + 8 ; where (x/a) = (p+q)

hence, y = (a/2)(p^2 + q^2) = (a/2)[x^2/a^2 + 8) ---> y = x^2/(2a) + 4a is the locus of the Midpoint of PQ.


Part (iii)

re-arranging the equation of the locus to: 2a(y -4a) = x^2 ---> (0, 4a) is the focus; and, (a/2) is the focal length.


hope that helps

 

[king_of_boredom]

wow, thanks who_loves maths.

i managed to get question B ok, but i think i got lost on question A. but it's still good to know that i can still get help on the holidays!