Locus Q. (1 Viewer)

[Smile12345]

Hello All.

t's a bit of a dumb question. But appreciate any/all help.

Q(2aq, aq^2) lies on the parabola x^2 = 4ay.

a)Find the equation of the focal chord through Q.
b)Prove that the length of the latus rectum is 4a.

Thanks in advance.

 

[HSC2014]

a. Use y-y1 = m(x-x1)
b. Use the fact that a point P(x,y) on a parabola is equidistant from the focus and directrix.

 

[Smile12345]

a. Use y-y1 = m(x-x1)
b. Use the fact on a point P(x,y) on a parabola is equidistant from the focus and directrix.

Yeah, ok then... Thanks.

 

[Smile12345]

Sorry can you please explain a further?

 

[HSC2014]

a: The focal chord is the chord that passes through the focus S(0,a) in this case
So you can find the gradient m of QS using y-y1/x-x1 (rise/run)
Then apply the formula y-y1 = m(x-x1) to find the equation of the chord

b: The latus rectum is the focal chord parallel to the directrix. In this case, the two extremities (ends) of the chord will be A(x1, a) and B(x2, a). Using the fact that a point on x^2 = 4ay is equidistant from the focus and directrix, we can see that x1 = +- 2 and x2 = -+ 2a (draw up a diagram if needed)

Hence distance AB (i.e. length of latus rectum) = 2a + 2a = 4a

 

[hit patel]

On any parabola there two points or more that can be taken into consideration. For a upwards x^2=4ay parabola, the points are as follows: (2ap,ap^2) and (2aq, aq^2).
a) Using gradient formula for these two points we get gradient = (p+q)/2
Now using point gradient formula: y-ap^2= [(p+q)/2](x-2ap)
Expanding the RHS and simplifying we y=1/2(p+q)x-apq

b) On x^2=4ay the focus is always (0,a)
Sub 0,a in focal chord found above to prove pq=-1 and therefore prove it to be focal chord. Now endpoints of focal chords are the points (2ap,ap^2) and (2aq, aq^2), therefore the distance can be found by using distance formula with these two coordinates and for sure you will find that this will equal to 4a. ( sorry didnt do distance formula working out since I donot know how to use sqrt signs on bos and will just make it more confusing).

Cheers

 

[Smile12345]

a: The focal chord is the chord that passes through the focus S(0,a) in this case
So you can find the gradient m of QS using y-y1/x-x1 (rise/run)
Then apply the formula y-y1 = m(x-x1) to find the equation of the chord

b: The latus rectum is the focal chord parallel to the directrix. In this case, the two extremities (ends) of the chord will be A(x1, a) and B(x2, a). Using the fact that a point on x^2 = 4ay is equidistant from the focus and directrix, we can see that x1 = +- 2 and x2 = -+ 2a (draw up a diagram if needed)

Hence distance AB (i.e. length of latus rectum) = 2a + 2a = 4a

Thanks... For the very detailed explanation... I do, what the parts are but not sure what do with them!

 

[Smile12345]

On any parabola there two points or more that can be taken into consideration. For a upwards x^2=4ay parabola, the points are as follows: (2ap,ap^2) and (2aq, aq^2).
a) Using gradient formula for these two points we get gradient = (p+q)/2
Now using point gradient formula: y-ap^2= [(p+q)/2](x-2ap)
Expanding the RHS and simplifying we y=1/2(p+q)x-apq

b) On x^2=4ay the focus is always (0,a)
Sub 0,a in focal chord found above to prove pq=-1 and therefore prove it to be focal chord. Now endpoints of focal chords are the points (2ap,ap^2) and (2aq, aq^2), therefore the distance can be found by using distance formula with these two coordinates and for sure you will find that this will equal to 4a. ( sorry didnt do distance formula working out since I donot know how to use sqrt signs on bos and will just make it more confusing).

Cheers

Thanks Man... (No worries about sqrt signs).

 

[HeroicPandas]

Therefore lactus rectum = 4a

 

[Smile12345]

View attachment 28611

Therefore lactus rectum = 4a

Yup, ok. thanks.

 

[HSC2014]

On any parabola there two points or more that can be taken into consideration. For a upwards x^2=4ay parabola, the points are as follows: (2ap,ap^2) and (2aq, aq^2).
a) Using gradient formula for these two points we get gradient = (p+q)/2
Now using point gradient formula: y-ap^2= [(p+q)/2](x-2ap)
Expanding the RHS and simplifying we y=1/2(p+q)x-apq

b) On x^2=4ay the focus is always (0,a)
Sub 0,a in focal chord found above to prove pq=-1 and therefore prove it to be focal chord. Now endpoints of focal chords are the points (2ap,ap^2) and (2aq, aq^2), therefore the distance can be found by using distance formula with these two coordinates and for sure you will find that this will equal to 4a. ( sorry didnt do distance formula working out since I donot know how to use sqrt signs on bos and will just make it more confusing).

Cheers

For a: Why did you introduce a second point P(2ap,ap^2)? I thought you would just focus on what you're given; point Q(2aq,aq^2) and point S(0,a). And two points is sufficient to find an equation

 

[Smile12345]

Sorry, can someone please explain a) fully? I'm just not getting it.

Thanks.