Quick locus questions (1 Viewer)

[atar90plus]

Hello

Just want to ask you guys to help me with a few problems I am currently having with locus. Please show full working and solution

5. d) Find the equation of the parabola with coordinates of the focus (0,2) and the coordinates of the vertex (0,0)
e) ^^^ with coordinates of the vertex (0,0) , equation of the axis x=0 and focal length 3
f) ^^^ with vertex (0,0) and the equation of the axis x=0 and passing through the point (-1,7)

7. Find the equation of the focal chord that cuts the curve x^2=8y at (-4,2). Please note for this question I just need help with finding the gradient coz for some reason I kept on getting the gradient as undefined

 

[Sy123]

Hello

Just want to ask you guys to help me with a few problems I am currently having with locus. Please show full working and solution

5. d) Find the equation of the parabola with coordinates of the focus (0,2) and the coordinates of the vertex (0,0)
e) ^^^ with coordinates of the vertex (0,0) , equation of the axis x=0 and focal length 3
f) ^^^ with vertex (0,0) and the equation of the axis x=0 and passing through the point (-1,7)

7. Find the equation of the focal chord that cuts the curve x^2=8y at (-4,2). Please note for this question I just need help with finding the gradient coz for some reason I kept on getting the gradient as undefined

5d)
Ignore this if you know why a parabola is a parabola
Cool way: We know by definition that a parabola is a locus of a variable point, where the distance from the variable point to a fixed point, and the variable point to any other variable point on a straight line is the same: This fixed point is our focus:

Take variable point F(x,y) and our fixed point (focus), as S(0,2). Now lets find the distance between F and S:

Now we need to find our fixed line (directix), but we are going to use the cool method:
Since the distance from our vertex to our focus, and the perpendicular distance from the vertex to our unknown directix should be the same:
Since the distance from vertex to focus is 2
Our directrix is y=-2 (perpendicular distance from vertex to this line is 2)

Okay now lets take the perpendicular distance of our variable F(x,y) and our line y=-2:



But because this equals to our original expression:



Now lets stop having fun and a apply the formula:

5e)
In our formula:

'a' is our focal length, in this it is 3, but also since our axis is x=0 it is a right side up parabola (a function)
Lets sub in a=3


Theoretically this also works:

5f)Right side up parabola with vertex zero:
We can do this:


Lets sub in a pair of values to get a: x=-1 y=7



Hence our equation is:



7) We know that its a focal chord, so the coordinates are S(0,a), but we know that a=2 due to the equation of the parabola (x^2=8y)
So we have two points now:
(0,2) and (-4,2)
Finding the gradient between them:

Into point-gradient formula:




That is the equation of the focal chord (this is a latus rectum since it is parallel to the x-axis), now it would also be good to notice from the very beginning that they have the same y value, hence any linear equation to go through these points must be y=2, since the y value is constant

 

[atar90plus]

Alright thanks mate

5d)
Ignore this if you know why a parabola is a parabola
Cool way: We know by definition that a parabola is a locus of a variable point, where the distance from the variable point to a fixed point, and the variable point to any other variable point on a straight line is the same: This fixed point is our focus:

Take variable point F(x,y) and our fixed point (focus), as S(0,2). Now lets find the distance between F and S:

Now we need to find our fixed line (directix), but we are going to use the cool method:
Since the distance from our vertex to our focus, and the perpendicular distance from the vertex to our unknown directix should be the same:
Since the distance from vertex to focus is 2
Our directrix is y=-2 (perpendicular distance from vertex to this line is 2)

Okay now lets take the perpendicular distance of our variable F(x,y) and our line y=-2:



But because this equals to our original expression:



Now lets stop having fun and a apply the formula:

5e)
In our formula:

'a' is our focal length, in this it is 3, but also since our axis is x=0 it is a right side up parabola (a function)
Lets sub in a=3


Theoretically this also works:

5f)Right side up parabola with vertex zero:
We can do this:


Lets sub in a pair of values to get a: x=-1 y=7



Hence our equation is:



7) We know that its a focal chord, so the coordinates are S(0,a), but we know that a=2 due to the equation of the parabola (x^2=8y)
So we have two points now:
(0,2) and (-4,2)
Finding the gradient between them:

Into point-gradient formula:




That is the equation of the focal chord (this is a latus rectum since it is parallel to the x-axis), now it would also be good to notice from the very beginning that they have the same y value, hence any linear equation to go through these points must be y=2, since the y value is constant