Some Prelim help, please? (1 Viewer)

[tarasomerville]

Firstly need some help with the topic Parabola and the Locus

- Find the equation of the locus of point P(x,y) that moves so that it is equidistant from the x axis and the u axis.

- find the equation of the locus of a point that moves so it's distance from the line 3x+4y+5=0 is always 4 units.

The last one is the first of a series of similar questions and I just need a little walk through on how to do these questions.

Thank you

 

[jeffwu95]

is the first one just y= +/- x
and is the second one just applying the perpendicular distance formula

 

[Sy123]

Firstly need some help with the topic Parabola and the Locus

- Find the equation of the locus of point P(x,y) that moves so that it is equidistant from the x axis and the u axis.

- find the equation of the locus of a point that moves so it's distance from the line 3x+4y+5=0 is always 4 units.

The last one is the first of a series of similar questions and I just need a little walk through on how to do these questions.

Thank you

These are the equations that satisfy the locus. What I did was I found the relation between fixed line 3x+4y+5=0, and a variable point (x,y), where the perpendicular distance between them was always 4. I subbed it in the Perpendicular distance formula, due to absolute value you get 2 Cartesian equations (as one would imagine).

EDIT:

Also the first one, it is y= x , but if you want to make it a single relation you can do it as


I dont think you will need to know that though.

 

[tarasomerville]

Thank you both.

More questions I can't figure out:

- find the equation of the locus of a point that moves so it is equidistant from the line 4x-3y+2=0 and 3x+4y-7=0.

Similarly,
- find the equation of the locus of a point that moves so it is equidistant from the line 3x+4y-5=0 and 5x+12y-1=0

Something to do with simultaneous equations I imagine?

 

[jeffwu95]

Thank you both.

More questions I can't figure out:

- find the equation of the locus of a point that moves so it is equidistant from the line 4x-3y+2=0 and 3x+4y-7=0.

Similarly,
- find the equation of the locus of a point that moves so it is equidistant from the line 3x+4y-5=0 and 5x+12y-1=0

Something to do with simultaneous equations I imagine?

for i) you let a point P (x1,y1) be the locus and do the perpindicular distance formula for both lines. ie distance=(4x1-3y1+2)/5 and distance=3x1+4y1-7/5 . you then equate the distances as it is equidistant and u get the locus of the point
ii) is exactly the same reasoning

 

[tarasomerville]

thank you

One more in this exercise:

If R is the fixed point (3,2) and P is a movable point (x,y) find the equation of the locus of P if the distance PR is twice the distance from P to the line y=-1

 

[jeffwu95]

you use the distance formula to calculate the distance between P and R ie sqrt[(3-x)^2+(2-y)^2]
then u calculate the perpendicular distance between P and the line y=-1
and as yknow the distance PR is twice the distance P to line y=-1 you let sqrt[(3-x)^2+(2-y)^2] = 2 times (.....)
sorry its late and i ceebs to calculate the distance between p and y=-1

 

[tarasomerville]

Thank you!

Some more.. Oh by the way these are between a friend and myself, only asking questions neither of us can get (or finish).

- Given two points A(2,-5) and B(-4,3), find the equation of the circle with diameter AB.
We got the length of AB which gave us diameter and radius, but we don't know how to get the answer.

 

[Amundies]

Find the midpoint of AB, that gives you the centre of the circle. Then just use the equation of a circle (x-h)^2 + (y-k)^2 = r^2, where h,k is the centre of the circle.

 

[Sy123]

To find the equation of the circle we need to find the radius and the centre of the circle, the radius is half the diameter and we are given the diameter, moreover the midpoint of the diameter is the centre of the circle. So lets find the two pieces of information:
Midpoint (M):



Therefore our centre is M(-1,-1).

Now lets find the radius which is the distacne MA:



Now we have radius=5 Centre(-1,-1). Now lets input into the eqn of the cirlce:

 

[tarasomerville]

Thanks, I actually did midpoint but didn't get as far as the equation.

- a circle has centre C(-1,3) and radius 5 units.
- equation: x^+2x+y^-6y-15=0
- the line 3x-y+1=0 meets the circle at two points. Find their coordinates. (abs values?)
- let coordinates be x and y, where y is the coordinate directly below the centre C. Find the coordinates of point z, where yz is a diameter of the circle.
- hence show angle zxy = 90'