could some1 explain this to me (1 Viewer)

[bos1234]

The Parametric eqn of the chord: y=1/2(p+q)x -apq

If the parameters p and q are changed, then the formulae fpr tje gradient of the chord amd tje equation of the chord remain the same. Geometrically this is because the chord PQ is the same line as the chord QP. Such expressions are called symmetric in p and q

what does the one in bold mean?

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P and Q are the pts with parameters p and q on the parabola x=2at,y=at^2

a) Chord PQ ==> y-1/2(p+q)x + apq = 0

b)chord when extended pass through the pt (0,-a) ==> show pq = 1

c) Hence, if S is the focus of the parabola, show that 1/SP + 1/SQ = 1/a

 

[ssglain]

Quite clearly, because p and q are both arbitrary third party parameters, they are perfectly interchangeable and it makes no difference what you use to represent either endpoints of a chord.

a) Use two-point formula.

b) Substitute (0, -a) into the equation for PQ.

c) Use distance formula to find SP = a*(p^2 + 1) and SQ = a*(q^2 + 1). Substitute everything into the RHS and stir around a bit.

RHS = 1/SP + 1/SQ = (SQ + SP)/SQ*SP
= [a*(p^2 + q^2 + 2)]/[a^2*(p^2 + 1)*(q^2 + 1)]
= (p^2 + q^2 + 2)/[a*(p^2*q^2 + p^2 + q^2 + 1)]
= (p^2 + q^2 + 2)/[a*(p^2 + q^2 + 2)] Since pq = 1 => p^2*q^2 = 1
= 1/a = LHS

 

[bos1234]

ok thanks i understand now

last one if not busy

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I done part a),b),c) but part d how can i do it?


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P and Q are the points t=p and t=q on the parabola

abd

a) Find the equations of the normals to the curve at P and Q

b)Prove that

c)Show that the normals intersect at at the point

d) If pq = 2, show that the normals intersect on the parabola



a) Parametric eqn for normal :

Pt P eqn

Pt Q eqn

 

[ssglain]

If pq=2 then the normals intersect at N[-2a(p + q), a(p^2 + q^2 + 2pq)]

Notice how if T = -(p + q), then T^2 = [-(p + q)]^2 = p^2 + q^2 + 2pq
N[-2a(p + q), a(p^2 + q^2 + 2pq)] can then be expressed as (2aT, aT^2) and this satisfies the parametric equation of the parabola.