Parametrics (1 Viewer)

[AnthonyO]

P is a point in the first quadrant of the co-ordinate plane, lying on the parabola x^2=4y. The normal to the parabola at P meets the parabola again at point Q in the second quadrant. The tangents to the Parabola at P and Q at a point T. If S is the focus of the parabola and if QS=2PS, show that:
i) QP subtends a right angle at S
ii) PQ=PT

Can anyone help out with this one? I didnt have dificulty drawing the graph, but i did with proving them.

 

[Robert1961]

X^2 = 4ay
X^2 = 4y therefore a = 1

Let P(2ap,ap^2) = P(2p,p^2) and Q(2q,q^2) (p>0, q<0)

Show eqn of tangent at P is y = px – p^2
Show eqn of tangent at Q is y = qx – q^2

Solve simultaneously to show T(p+q , pq)

Show eqn of normal of normal at P is y – p^2 = -(1/p)(x – 2p)
Show eqn of normal of normal at P is y = -x/p + 2 + p^2

Solve simultaneously with y = (x^2)/4

To get px^2 + 4x -8p – 4p^3 = 0

Use quadratic formula to show
x = 2p (which is abscissa of P) or x= (-4 – 2p^2)/p (which is abscissa of Q)

Using y = x^2/4 ShowQ( (-4 – 2p^2)/p , ((2 + p^2)^2)/(p^2) )

Given QS = 2PS
Then QS^2 = 4PS^2

Some painful distance formulas squared and put LHS over a common denominator of p^4
Then multlipy both sides by p^4

Arrives at
p^8 + 10p^6 + 33p^4 + 40p^2 + 16 = 4p^8 + 8p^6 + 4p^4

0 = 3p^8 – 2p^6 – 29p^4 – 40p^2 – 16

try p= 2 (LHS = RHS) therefore p = 2 (due to the nature of the signs of the coefficients & the powers all being even you can argue that no other positive solution for p exists)

therefore P(4 ,4) & Q(-6 , 9) hence q is -3

Consider grad PS X grad QS
= ((4-1)/4) X ((9-1)/-6)
= -1

Therefore PS perpendicular to QS (Part (i) QED)

Sub p = 2 & q = -3 into T((p + q) , pq) then T(-1 , -6)

PQ = SQRT((4 - -6)^2 + (4 – 9)^2) = SQRT(125)
PT = SQRT((4 - -1)^2 + (4 - -6)^2) = SQRT(125)

Therefore PQ = PT (both = SQRT(125)) (Part (ii) QED)

I will be checking back to this thread to see if others come up with a much simplier solution.

 

[jyu]

Using y = x^2/4 ShowQ( (-4 – 2p^2)/p , ((2 + p^2)^2)/(p^2) )

Given QS = 2PS
Then QS^2 = 4PS^2

Some painful distance formulas squared and put LHS over a common denominator of p^4
Then multlipy both sides by p^4

Arrives at
p^8 + 10p^6 + 33p^4 + 40p^2 + 16 = 4p^8 + 8p^6 + 4p^4

0 = 3p^8 – 2p^6 – 29p^4 – 40p^2 – 16

try p= 2 (LHS = RHS) therefore p = 2 (due to the nature of the signs of the coefficients & the powers all being even you can argue that no other positive solution for p exists)

 

[Robert1961]

Thx Jyu - I new there had to be a way of avoiding a polynomial of deg 8

Good work that man

 

[Robert1961]

Hi Jyu,

(p - 1/p)(p - 4/p) = 0 should be

(p + 1/p)(p - 4/p) = 0

Also a square is missing on the 2nd pair or square brackets on the line labelled ....(1)

 

[Robert1961]

Hi AnthonyO,

where did that question come from?

 

[cl3nta]

nice question

 

[jyu]

Hi Jyu,

(p - 1/p)(p - 4/p) = 0 should be

(p + 1/p)(p - 4/p) = 0

Also a square is missing on the 2nd pair or square brackets on the line labelled ....(1)

Thanks for pointing out the error.