Easy Parametrics Q (1 Viewer)

[azureus88]

"Find the coordinates of 3 points on the parabola x^2 = 4ay such that the normals through these 3 points pass through the point (-12,15)"

Is there a better way than forming 3 equations in 3 unknowns?

 

[gurmies]

quite a unique question. I think I would just do what you said, 3 equations in 3 unknowns. Using parameters like, p, q, r. I think though the values will be in terms of a?

 

[Timothy.Siu]

"Find the coordinates of 3 points on the parabola x^2 = 4ay such that the normals through these 3 points pass through the point (-12,15)"

Is there a better way than forming 3 equations in 3 unknowns?

i've done this question around 10 times now, but the question isn't like that as well, if its the fitzpatrick question, it's through x^2=4y not x^2=4ay
theres a few ways u can do this, u might not like my way but here it is

gradient of tangent y'=x/2
gradient of normal -2/x
equation of normal passing through (-12,15)
y-15=(-2/x)(x+12)
y=-2-24/x+15
to find points, we find the intersection of this and equation of parabola y=x^2/4
we get, -2-24/x+15=x^2/4 multiply both sides by 4x and move to one side
x^3-52x+96=0
then we can solve this polynomial,x=2,x=6 x=-8 and find the y values and ur done

 

[azureus88]

i've done this question around 10 times now, but the question isn't like that as well, if its the fitzpatrick question, it's through x^2=4y not x^2=4ay
theres a few ways u can do this, u might not like my way but here it is

gradient of tangent y'=x/2
gradient of normal -2/x
equation of normal passing through (-12,15)
y-15=(-2/x)(x+12)
y=-2-24/x+15
to find points, we find the intersection of this and equation of parabola y=x^2/4
we get, -2-24/x+15=x^2/4 multiply both sides by 4x and move to one side
x^3-52x+96=0
then we can solve this polynomial,x=2,x=6 x=-8 and find the y values and ur done

nah, i reckon thats a pretty efficient method. thanks. btw can u post a few good parametrics questions u've come across, so i can practice with?

 

[gurmies]

Very nice method there

 

[Fortian09]

Hmm i might just add some more parametrics questions (well at least i think they're parametrics questions coz the stupid sheet just ses "Parabola:... ><')

Okay first one

Find the equation of the tangent at the point P(8,2) on the parabola x²=32y. If this tangent meets the tangent at the vertex of the parabola at Q, find the coordinates of Q.
( I got the first part in this question, it's the second part thats gotten me stumped)

Find the equation of the directrix of the parabola x²=-12y. The equation of the tangent at the point (-6,-3) on this parabola meets the directrix at T. Find the coordinates of T.

P(-2,1) and Q(6,9) are points on the parabola x²=4y.
The line through M, the midpoint of PQ, parallel to the axis of the parabola meets the parabola in N.
i) Find the coordinates of M and N.
ii) Show that the tangent at N is parallel to the chord PQ.

 

[azureus88]

Hmm i might just add some more parametrics questions (well at least i think they're parametrics questions coz the stupid sheet just ses "Parabola:... ><')

Okay first one

Find the equation of the tangent at the point P(8,2) on the parabola x{sup]2[/sup]=32y. If this tangent meets the tangent at the vertex of the parabola at Q, find the coordinates of Q.
( I got the first part in this question, it's the second part thats gotten me stumped)

Find the equation of the directrix of the parabola x²=-12y. The equation of the tangent at the point (-6,-3) on this parabola meets the directrix at T. Find the coordinates of T.

P(-2,1) and Q(6,9) are points on the parabola x²=4y.
The line through M, the midpoint of PQ, parallel to the axis of the parabola meets the parabola in N.
i) Find the coordinates of M and N.
ii) Show that the tangent at N is parallel to the chord PQ.

well, the tangent at the vertex of the parabola is y=0 (ie the x axis) so just sub y=0 into the equation of tangent at P. This will give you the x-coordinate.

 

[Fortian09]

Thats coz it meets at the tangent right?

what about the other two?

 

[12o9]

Hmm i might just add some more parametrics questions (well at least i think they're parametrics questions coz the stupid sheet just ses "Parabola:... ><')

Okay first one

Find the equation of the tangent at the point P(8,2) on the parabola x²=32y. If this tangent meets the tangent at the vertex of the parabola at Q, find the coordinates of Q.
( I got the first part in this question, it's the second part thats gotten me stumped)

Find the equation of the directrix of the parabola x²=-12y. The equation of the tangent at the point (-6,-3) on this parabola meets the directrix at T. Find the coordinates of T.

P(-2,1) and Q(6,9) are points on the parabola x²=4y.
The line through M, the midpoint of PQ, parallel to the axis of the parabola meets the parabola in N.
i) Find the coordinates of M and N.
ii) Show that the tangent at N is parallel to the chord PQ.

 

[Fortian09]

problems related to Locus not sure if it fits into parametrics though...


1. The area o the triangle enclosed by a variable line L and the axes is 8square units. M is the midpoint of the line segment on L cutting by the axes. Find the equation of the locus of M.

2. A point A moves on the line 5x-y+3=0 while a point B moves on the curve x²=y+3, and AB is parallel to the y-axis. Find the equation of the locus of the midpoint M of AB.

3. A line has slope m and y-intercept m+2. Another line has slope m+1 and y-intercept 1. Find the locus of the intercepting point of the two lines as m varies.

 

[azureus88]

problems related to Locus not sure if it fits into parametrics though...


1. The area o the triangle enclosed by a variable line L and the axes is 8square units. M is the midpoint of the line segment on L cutting by the axes. Find the equation of the locus of M.

2. A point A moves on the line 5x-y+3=0 while a point B moves on the curve x²=y+3, and AB is parallel to the y-axis. Find the equation of the locus of the midpoint M of AB.

3. A line has slope m and y-intercept m+2. Another line has slope m+1 and y-intercept 1. Find the locus of the intercepting point of the two lines as m varies.

ok, i got the following answers (which are probably wrong):

1. xy=1
2. y=(x^2 + 5x)/2
3. y=x^2 +1

 

[Fortian09]

i have the answers but i actually need the worked solutions sorry

 

[azureus88]

i have the answers but i actually need the worked solutions sorry

well, are they right first? cause i dont wanna post wrong solutions

 

[Fortian09]

i dun care about the answer i just want the steps right

 

[clintmyster]

i dun care about the answer i just want the steps right

like azureus said, if you want correct solutions, inform us of the answers first!

 

[Trebla]

problems related to Locus not sure if it fits into parametrics though...


1. The area o the triangle enclosed by a variable line L and the axes is 8square units. M is the midpoint of the line segment on L cutting by the axes. Find the equation of the locus of M.

2. A point A moves on the line 5x-y+3=0 while a point B moves on the curve x²=y+3, and AB is parallel to the y-axis. Find the equation of the locus of the midpoint M of AB.

3. A line has slope m and y-intercept m+2. Another line has slope m+1 and y-intercept 1. Find the locus of the intercepting point of the two lines as m varies.

1) Let the intercepts of the line L be (x₁,0) and (0,y₁)
x₁y₁ / 2 = 8
x₁y₁ = 16
Midpoint of L is M (x₁ / 2,y₁ / 2)
So locus of M is with x₁ and y₁ parameters:
x = x₁ / 2
y = y₁ / 2
xy = x₁y₁ / 4
.: xy = 4

2) Let A be (x₁, y₁) and B be (x₁, y₂) as AB is parallel to the y-axis, so
5x₁ - y₁ + 3 = 0
x₁² - y₂ - 3 = 0
Adding the above equations:
5x₁ + x₁² - y₁ - y₂ = 0
Midpoint of AB is M(x₁, [y₁ + y₂] / 2)
Locus of M is:
x = x₁
y = (y₁ + y₂) / 2
=> 2y = y₁ + y₂
Using:
5x₁ + x₁² - y₁ - y₂ = 0
5x + x² - 2y = 0

3) The equation of the lines are y = mx + m + 2 and y = (m + 1)x + 1
Finding intersecting points:
mx + m + 2 = mx + x + 1
=> x = m + 1
Sub into y = (m + 1)x + 1
.: y = x² + 1

 

[Fortian09]

Thanks for some useful help

I have a problem constructing parametric equations... >< :$
and a few cartesian equations.

1. x=t²-2t, y=t²+2

2. x=t²+t, y=(t²/2) - 3t

3. x=3[t+(1/t)], y=4[t-(1/t)]

4. x=1/2 (a^t+a^-t), y=1/2 (a^t-a^-t)

5. x=7(sec@+tan@), y=7(sec@-tan@)

6. x=t+ 1/t, y=t- 1/t

7. 2x-y+4=0

8. x[/sup]2[/sup]+y²=9



A=(5,-4), B = (-3,2). Find the locus of P if the point P(x,y) move such that the distance of P from the line AB equal s distance of P from x=5.

 

[azureus88]

To construct cartesian equations from parametric equations, you gotta try to eliminate the parameter.

1. x=t^2 - 2t (1)
y= t^2 + 2 (2)

y-x = 2+2t
so t = (y-x-2)/2
substitute this into eqn (1) or (2)

2. x=t<SUP>2</SUP>+t
y=(t<SUP>2</SUP>/2) - 3t
2y=t^2 - 6t

x-2y = 7t
so t = (x-2y)/7
substitute this into initial equation

3. x=3[t+(1/t)]
y=4[t-(1/t)]

x/3 = t+(1/t)
y/4 = t-(1/t)

[t+(1/t)]^2 = [t-(1/t)]^2 + 4
so x^2/9 - y^2/16 = 4



the rest follow similar procedures. for 5, use the identity sec^2@ - tan^2@ =1

 

[Fortian09]

wt about 7 nd 8 when it's the reverse?

 

[Timothy.Siu]

7. 2x-y+4=0

x=t-2 y=2t

8. x²+y<SUP>2</SUP>=9

x=t y=sqareroot(9-t²)

i think...
u can just sub in random stuff