Parametrics (1 Viewer)

[mtsmahia]

Hi guys once again,

can some help with this please

P is a variable point on the parabola x=2t, y=t^2 with focus S and vertex A. Q is the midpoint of SP and R the midpoint of AQ. Find the cartesian equation of the locus of R. Show that the locus is a parabola and find the vertex and focus.

thanks

 

[ninetypercent]

a = 1/2
the parabola x^2 = 4ay
x^2 = 2y

focus (0,a) => S (0,1/2)
vertex (0,0)
P (2ap,ap^2) => (p, p^2/2)

Q = [p/2, (p^2 + 1)/4]

R is the midpoint of AQ

R (p/4, (p^2 + 1)/8)

x = p/4 => 4x = p
y = (p^2 + 1)/8
= (16x^2 + 1)/8

8y = 16x^2 +1

it is a parabola

 

[martinc]

P is a variable point on the parabola x=2t, y=t^2 with focus S and vertex A. Q is the midpoint of SP and R the midpoint of AQ. Find the cartesian equation of the locus of R. Show that the locus is a parabola and find the vertex and focus.

Parabola is x^2 = 4y , and then a =1

p(2p,p^2)
focus (0,1)

Q ( p, [p^2+1] / 2)

R ( p/2 , [p^2+1] / 4)

for locus R

x = p/2

y p^2+1 /4

then make p the subject, and sub it into y

p=2x

therefore locus of R is y = [4x^2 +1] /4

4y = 4x^2 +1
4x^2 = 4y -1
x^2 = y-1/4

edit:LOL beat me to it by a long way.

 

[mtsmahia]

answer is x^2=2(y- 1/2)

 

[math man]

P is a variable point on the parabola x=2t, y=t^2 with focus S and vertex A. Q is the midpoint of SP and R the midpoint of AQ. Find the cartesian equation of the locus of R. Show that the locus is a parabola and find the vertex and focus.

Parabola is x^2 = 4y , and then a =1

p(2p,p^2)
focus (0,1)

Q ( p, [p^2+1] / 2)

R ( p/2 , [p^2+1] / 4)

for locus R

x = p/2

y p^2+1 /4

then make p the subject, and sub it into y

p=2x

therefore locus of R is y = [4x^2 +1] /4

4y = 4x^2 +1
4x^2 = 4y -1
x^2 = y-1/4

edit:LOL beat me to it by a long way.

i worked it out and i agree with this answer, therefore vertex is at (0, 1/4) and focus at (0, 1/2)