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Sholpan [36]
3 years ago
13

at a point P on the parabola x^2=4ay a normal PK is drawn. From vertex O, a perpendicular OM is drawn to meet the normal at M. S

how that the equation of the locus of M as P varies on the parabola is x^4-2ax^2y+x^2y^2-ay^3=0
Mathematics
1 answer:
saveliy_v [14]3 years ago
6 0
<span>Answer: Its too long to write here, so I will just state what I did. I let P=(2ap,ap^2) and Q=(2aq,aq^2) But x-coordinates of P and Q differ by (2a) So P=(2ap,ap^2) BUT Q=(2ap - 2a, aq^2) So Q=(2a(p-1), aq^2) which means, 2aq = 2a(p-1) therefore, q=p-1 then I subbed that value of q in aq^2 so Q=(2a(p-1), a(p-1)^2) and P=(2ap,ap^2) Using these two values, I found the midpoint which was: M=( a(2p-1), [a(2p^2 - 2p + 1)]/2 ) then x = a(2p-1) rearranging to make p the subject p= (x+a)/2a</span>
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You can check this answer by adding up all of the parts we used in this problem:

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