2 Unit Locus Questions (1 Viewer)

[V_L]

Hi

Can someone please go through how to answer questions 24 and 25?

The answers are also attatched.

Thank you !!

 

[nrumble42]

Q24.

SO the gradient of PA is y + 2/x-3, simply by using the formula.
similarly, the gradient of PB is y-7/x +1

Now gradient PA=2PB

therefore, y+2/x-3 = 2(y-7)(x+1). Now cross multiply the demoninators and you'll get:

(y + 2)(x+1)= 2(y-7)(x-3). Expand this and you'll get:

xy + y +2x +2= 2(xy-3y-7x+ 21)

which ends up as: 0=xy-7y-16x+40

 

[pikachu975]

Hi

Can someone please go through how to answer questions 24 and 25?

The answers are also attatched.

Thank you !!

Question 24
Gradient of PA:
m1 = (y+2)/(x-3)

Gradient of PB:
m2 = (y-7)/(x+1)

m1 = 2*m2
(y+2)/(x-3) = 2(y-7)/(x+1)
(y+2)(x+1) = (2y-14)(x-3)
xy + y + 2x + 2 = 2xy - 6y - 14x + 42
xy - 7y - 16x + 40 = 0 QED

Question 25
Distance of PR:
d^2 = (x-3)^2 + (y-2)^2
d = square root of that

Distance of P to y=-1:
d^2 = (x-x)^2 + (y+1)^2
d^2 = (y+1)^2
d = y + 1

PR = 2*(P to y=-1)
sqrt[ (x-3)^2 + (y-2)^2 ] = 2(y+1)
(x-3)^2 + (y-2)^2 = 4(y+1)^2 ... (squaring both sides)
x^2 - 6x + 9 + y^2 - 4y + 4 = 4y^2 + 8y + 4
x^2 - 6x - 3y^2 - 12y + 9 = 0

 

[nrumble42]

Note that in Q25, to to the distance of the point P to the line y=-1, you can also do the perpendicular distance formula as that is quicker

 

[InteGrand]

In fact, since the line y = -1 is a line of the form y = const., we can immediately write down the distance of P(x, y) to this line as being |y+1| without need for calculation. (In general, distance of P(x, y) to the line y = C would be |y – C|.)

 

[InteGrand]

Note that for the second question, it is a hyperbola (with eccentricity 2), and its equation can be written in the form ((x – h)^2)/(A^2) – ((y – k)^2)/(B^2) = 1, where (h, k) is where the hyperbola is "centred". Since 4U students study conics, they should be able to get this answer and bypass a lot of algebra required in the 2U method.

 

[highshill]

Q24.

SO the gradient of PA is y + 2/x-3, simply by using the formula.
similarly, the gradient of PB is y-7/x +1

Now gradient PA=2PB

therefore, y+2/x-3 = 2(y-7)(x+1). Now cross multiply the demoninators and you'll get:

(y + 2)(x+1)= 2(y-7)(x-3). Expand this and you'll get:

xy + y +2x +2= 2(xy-3y-7x+ 21)

which ends up as: 0=xy-7y-16x+40

Why isn't it 2PA=PB

 

[Cena123]

Why isn't it 2PA=PB

Gradient of PA is twice PB
You need two lots of PB to get PA. Hence 2PB=PA