Question #4931b

1 Answer
Apr 17, 2017

#x= 2/9 ysqrt(36-y^2)#

Explanation:

The secret lies in using trig identities: #sin(2t) = 2sin(t)cost(t); sin(t) = sqrt(1-cos^2(t)). #
We can write #cos(t) = y/6; and x = 4*2sin(t)cos(t) .#
That is, #x = 8cos(t)sqrt(1-cos^2(t)); # than replace #cos(t)#, so #x = 8*y/6sqrt(1-(y/6)^2) =8/36ysqrt(36-y^2) = 2/9ysqrt(36-y^2).#