(Tech free)Determine the locus of points such that if P=(x,y) ,A=(-2,0) and B=(2,0) the PA+PB=4
Help! My exam is in 3 days!
Disclaimer: No idea if this working out is VCE friendly. Just thought I'd lend a hand.
Hey Gogo! We'll take the points and substitute them into expressions for the distances PA and PB to start:
Now let's play with that relationship a bit so we can bring in the expressions without the square roots:
Notice we got a whole heap of cancellation into that last line. That's what we wanted to see. We can continue:
So this locus lies on the x-axis; however, it isn't the whole x-axis. We need to check something else to find the missing restriction, remember that for our condition to make sense, PA must be less than or equal to 4, and so must PB. Let's check that condition:
These two regions overlap
only for \(-2\le x\le2\), so that's where our locus lies. Putting it together, this locus is just the line segment joining A and B.
Note that you could do this last bit just with some clever thinking. Clearly we can't venture beyond A or B with our locus, because that just doesn't make sense. As soon as we do, one of our distances is immediately, clearly, too large. You could probably not do any algebra for this question if you are good at picturing loci, but I thought I'd show everything
sorry if this isn't VCE friendly, but I thought I'd lend a hand anyway!