equation of the chord to parabola (1 Viewer)

[ozidolroks]

can someone please help me with this question???

a secant which passes through the point ( 0,-a), cuts the parabola x2 = 4ay at P, Q with parameters p,q

If S is the focus show that 1 / PS + 1/ QS= 1/a

 

[Timothy.Siu]

lol i got owned,
hmm must've done something wrong lol

ahh i got it i think

well, p(2ap, ap^2) q(2aq,aq^2) S(0,a)

gradient AP=(p^2+1)/2p
gradient AQ=(q^2+1)/2q
but gradient AP=gradient AQ
therefore (after some rearranging) pq=1

then Distance(PS)=sqrt[(2ap)²+(ap²-a)²)]
similar for distance QS except p's are replaced with q's
D(ps)=a(p²+1) after some simplifying
then D(qs)=a(q²+1)

q=1/p D(QS)=a(1/p²+1)=a(p²+1)/p²

then, LHS=1/a(p²+1) + p²/a(p²+1)=1/a=RHS

 

[azureus88]

equation of PQ (using 2 point formula) is:
y = (p+q)x - apq

it passes through (0, -a) so subbing it in equation gives:
-a = 0 -apq
pq = 1

to find PS and QS, you can either use distance formula or use definition of parabola (ie, distance from focus to point on parabola is equal to perpendicular distance from that point to the directrix y=-a)

using the 2nd method (which is quicker) gives:
PS = ap^2 + a
QS = aq^2 + a

1/PS + 1/QS = 1/[a(p^2+1)] + 1/[a(q^2+1)]
= 1/[a(p^2+1)] + 1/[a((1/p^2)+1)] since pq = 1 so q=1/p
=1/[a(p^2+1)] + (p^2)//[a(p^2+1)]
=(1+p^2)/[a(p^2+1)]
= 1/a

 

[Trebla]

can someone please help me with this question???

a secant which passes through the point ( 0,-a), cuts the parabola x2 = 4ay at P, Q with parameters p,q

If S is the focus show that 1 / PS + 1/ QS= 1/a

Let P be (2ap, ap²) and Q be (2aq, aq²)
Gradient of PQ:
m_PQ = (ap² - aq²) / (2ap - 2aq)
= a(p - q)(p + q) / 2a(p - q)
= (p + q) / 2
Equation of PQ:
y - ap² = (p + q)(x - 2ap) / 2
This line passes through (0, - a) so subbing it in:
- a - ap² = (p + q)(- 2ap) / 2
1 + p² = p(p + q)
1 + p² = p² + pq
=> pq = 1

With S(0, a):
PS = √[(2ap)² + (ap² - a)²]
= √[4a²p² + a²p⁴ - 2a²p² + a²]
= √[4a²p² + a²p⁴ - 2a²p² + a²]
= √[a²p⁴ + 2a²p² + a²]
= √(ap² + a)²
= a(p² + 1)
[Technically its |a(p² + 1)| but since a(p² + 1) always positive with a > 0 and (p² + 1) > 0, we can remove the absolute values]

Similarly QS = a(q² + 1)

.: 1 / PS + 1 / QS = 1 / a(p² + 1) + 1 / a(q² + 1)
= (q² + 1 + p² + 1) / a(p² + 1)(q² + 1)
= (p² + q² + 2) / a(p²q² + p² + q² + 1)
But pq = 1
= (p² + q² + 2) / a(1 + p² + q² + 1)
= (p² + q² + 2) / a(p² + q² + 2)
= 1 / a

 

[ozidolroks]

thanks... how do you know what condition for PQ in the parabola x^2= 4ay is to be perpendicular to the axis of the parabola ?

 

[Timothy.Siu]

thanks... how do you know what condition for PQ in the parabola x^2= 4ay is to be perpendicular to the axis of the parabola ?

if P and Q have the same y coordinates then PQ will be perpendicular to the axis for the parabola x^2=4ay
(i'm assuming this is not related to the initial question)

 

[ozidolroks]

nope ...see there is this question that says for the parabola x^2 = 4ay
a) show that the gradient of the chord joining the vertex to a point P (2ap, ap^2) is p/2.
b) If Q is the point (2aq,aq^2) state the gradient of OQ and hence show that the chords of OP, OQ are perpendicular if pq = -4.
c) what is the condition for PQ to be perpendicular to the axis of the parabola ?

i can do the first 2 questions i just don't understand c ) the answer is supposedly p = -q

 

[Trebla]

Timothy.Siu is correct.
The ordinates of P and Q have to be equal so:
ap² = aq²
p² - q² = 0
a(p - q)(p + q) = 0
.: p = q or p = -q
BUT p = q means that P and Q are the same point, if they are distinct points then they must lie on opposite sides of the parabola so p = -q

 

[Timothy.Siu]

oh i get it,

u see, P(2ap, ap^2) and Q(2aq, aq^2)

for PQ to be perpendicular to the axis, the y values will be the same and x values will be the same except one is positive and one is negative...
soo,
ap^2=aq^2
2ap=-2aq p=-q

u can just think about it though, if u draw a diagram, for something to be perpendicular to the axis, it has to be a horizontal line,

 

[ozidolroks]

this might sound like a really dumb question but you said that you draw a horizontal line to the axis, is that to the x or y axis ???

 

[Timothy.Siu]

this might sound like a really dumb question but you said that you draw a horizontal line to the axis, is that to the x or y axis ???

umm, well to draw a perpendicular to the axis of the parabola, it has the make a right angle with it so, it would be a line parallel to the x-axis

 

[ozidolroks]

thank you