Parabola - Parametric to Cartesian Conversion (2 more) (1 Viewer)

[gcalway]

Hi,

I had been progressing well with my understanding of parametrics until these two questions arose. The questions in full are:

1) Tangents to the parabola x^2 = 4ay at the points P(2ap,ap^2)and Q(2aq,aq^2) meet at R.
a) Show that the equation of the tangent at P is y - px + ap^2 = 0
b) Show that R is the point (a(a+p),apq)
c) Find the Cartesian equation of the locus of R given the following:
(i) pq = -2 (ii) PQ has gradient 3 (iii) q = p + 2

2) The normals to the parabola x^2 = 4ay at the points P(2ap,ap^2)and Q(2aq,aq^2) meet at R.
a) Show that the equation of the normal at P is py + x = 2ap + ap^3
b) Show that R is the point (-apq(p+q),a(p^2+q^2+pq+2)
c) Find the Cartesian equation of the locus of R given the following:
(i) pq = -2 (ii) PQ has gradient 3

I have been able to complete parts a) & b) for both questions but don't seem to understand parts c). I am unsure whether the question is looking for multiple or single answer for c) and my efforts so far (assuming all the givens apply at once) are:

1 c) y = -x/3 But the model I created suggest a near vertical line with a positive x-intercept

2 c) y = -10x/3 But the model suggests a line with positive gradient and positive y-intercept

I would appreciate assistance to understand how I should approach parts c) and also the answers.

Thanks in advance,

Graeme

 

[InteGrand]

Hi,

I had been progressing well with my understanding of parametrics until these two questions arose. The questions in full are:

1) Tangents to the parabola x^2 = 4ay at the points P(2ap,ap^2)and Q(2aq,aq^2) meet at R.
a) Show that the equation of the tangent at P is y - px + ap^2 = 0
b) Show that R is the point (a(a+p),apq)
c) Find the Cartesian equation of the locus of R given the following:
(i) pq = -2 (ii) PQ has gradient 3 (iii) q = p + 2

2) The normals to the parabola x^2 = 4ay at the points P(2ap,ap^2)and Q(2aq,aq^2) meet at R.
a) Show that the equation of the normal at P is py + x = 2ap + ap^3
b) Show that R is the point (-apq(p+q),a(p^2+q^2+pq+2)
c) Find the Cartesian equation of the locus of R given the following:
(i) pq = -2 (ii) PQ has gradient 3

I have been able to complete parts a) & b) for both questions but don't seem to understand parts c). I am unsure whether the question is looking for multiple or single answer for c) and my efforts so far (assuming all the givens apply at once) are:

1 c) y = -x/3 But the model I created suggest a near vertical line with a positive x-intercept

2 c) y = -10x/3 But the model suggests a line with positive gradient and positive y-intercept

I would appreciate assistance to understand how I should approach parts c) and also the answers.

Thanks in advance,

Graeme

Those part c)'s are asking for separate answers for each part. The conditions don't hold at once, they are separate Q's.

 

[gcalway]

Thanks Integrand for the quick response.

I thought that might be the case as in 1ci) and 1ciii) I could not get real values when solving for p & q simultaneously between (i) & (iii). But I thought I would seek expert opinion before continuing.

I will now return to the problem and see how I go solving them all individually.

Regards,

Graeme

 

[gcalway]

I would really appreciate it if someone could review my answers for 1c) and 2c). I think i have taken the correct approaches but would like to be sure.

1 c) (i) Convert R(a(p+q),apq) to Cartesian where pq = -2

Sub for pq in y = apq => y = -2a

(ii) Convert R(a(a+p),apq) to Cartesian where PQ has gradient 3

Gradient PQ = 3 = (Py - Qy)/(Px - Qx)

3 = (ap^2 - aq^2)/(2ap - 2aq)

3 = [a(p+q)(p-a)]/[2a(p-q)]

6 = (p+q)

Sub for (p+q) in x = a(p+q) => x = 6a

(iii) Convert R(a(a+p),apq) to Cartesian where q = p+2

Sub for q in x = a(p+q)

x = a(p + p+2)

x = a(2p + 2)

x = 2a(p+1)

x/2a = p + 1

p = (x - 2a)/2a

Sub for p & q in y = apq

y = ap(p+2)

y = a[(x-2a)/2a][(x-2a)/2a + 2]

y = a(x-2a)(x-2a+4a)/4a^2

4ay = x^2 - 2ax - 2x - 2ax - 4a^2 - 4a

4ay = x^2 - 4a(x+1) - 4a^2


2 c) (i) Convert R(-apq(p+q),a(p^2 + q^2 + pq + 2)) to Cartesian where pq = -2

Sub for pq in x = -apq(p+q) => x = +2a(p+q) => p+q = x/2a

Sub for pq in y = a(p^2 + q^2 +pq +2

y = a(p^2 + q^2 -2 + 2)

Completing the square gives

y = a(p^2 + 2pq + q^2) - 2apq

y = a(p+q)(p+q) + 4a

Sub for p+q y = a(x/2a)^2 + 4a

y = ax^2/4a^2 + 4a

y = (ax^2 + 16a^3)/4a^2 (divide through by 'a')

4ay = x^2 + (4a)^2

(ii) Convert R(-apq(p+q),a(p^2 + q^2 +pq +2)) to Cartesian where PQ has gradient 3

Gradient PQ = 3 = (Py - Qy)/(Px - Qx)

3 = (ap^2 - aq^2)/(2ap - 2aq)

3 = [a(p+q)(p-a)]/[2a(p-q)]

6 = (p+q)

Use completing the square to modify y = a(p^2 + q^2 +pq +2)

y = a(p^2 + 2pq + q^2 -2pq + pq + 2)

y = a[(p+q)(p+q) - pq + 2)

Sub for (p+q) = 6

y = 36 - apq + 2a

From x = -apq(p+q) we get x = -6apq => pq = -x/6a

Sub for pq

y = 36 - a(-x/6a) + 2a

y = x/6 + 36 + 2a


Thanks in advance,

Graeme