Another Paramtric Parabola questions (1 Viewer)

[Jackson94]

P and Q are the points with parameters p and q on the parabola x = 2at and y = at^2
If PQ is a focal chord (hence pq=-1) show that PQ=a(p + 1/p)^2
Thanks guys!

 

[bleakarcher]

P(2ap,ap^2), Q(2aq,aq^2). By distance formula:
=>PQ=sqrt[(2ap-2aq)^2+(ap^2-aq^2)^2]=sqrt[4a^2(p-q)^2+a^2(p-q)^2(p+q)^2]
But pq=-1=>q=-1/p
:.PQ=sqrt[a^2(p-q)^2[4+(p+q)^2]]=sqrt[a^2(p+(1/p))^2[4+(p-(1/p))^2]]=sqrt[a^2(p+(1/p))^4]
i.e. PQ=a[p+(1/p)]^2

 

[cssftw]

P and Q are the points with parameters p and q on the parabola x = 2at and y = at^2
If PQ is a focal chord (hence pq=-1) show that PQ=a(p + 1/p)^2
Thanks guys!

easier way mate:

so Q is the point (2aq, aq^2), and P is the point (2ap, ap^2)

Remember the locus of parabola is: set of points where distance from a line (directrix) is equal to distance from focus
If you look at the image, u can see SQ=QN AND PS=PM - following the rule of the locus of the parabola

So in order to calculate PQ, you don't need the distance formula, you can simply calculate QN + PM (as QN = SQ and PM=PS..... and PS+SQ = PQ)

the y-coordinate of Q is (aq^2).
the y-coordinate of N is (-a)
therefore QN = aq^2 + a = a(q^2 + 1)

Similarly, PM = a(p^2 + 1)

QN + PM = a(q^2 + 1) + a(p^2 + 1)
= a(p^2 + q^2 + 2)

Now you know pq = -1....therefore q=-1/p --> q^2 = 1/p^2

therefore

QN + PM = a(p^2 + 2 + (1/p)^2)
= a(p+(1/p))^2

Easier method, because it's an extremely easy concept, just remember with any focal chord and distance related questions, distance from the point on a parabola to its focus, is equal to the distance of that point to the directrix.


**EDIT***: If you think I helped/assisted you in the slightest, REP ME BRAH

 

[Jackson94]

Thanks mate