Question

Question #0d91d

Answer 1

`y= (3/4)(2-x^2).`

Explanation:

Recall the identity : `sin^2theta=(1-cos2theta)/2.`

Hence,
`y=3sin^2theta=(3/2)(1-cos2theta).`..............(1)

But, it is given that `x=sqrt(2cos2theta),`
so that `x^2/2=cos2theta.`

Now, putting this value of `cos2theta` in (1), we get,

`y= (3/2)(1-x^2/2)=(3/4)(2-x^2).`

Answer 2

`y=(x^2-2)/-2`

Explanation:

`y=3sin^2theta`
`x=sqrt(2cos2theta)`
`x^2=2cos2theta`
=`2cos^2theta-2sin^2theta`
=`2cos^2theta-2/3y`
=`2(1-1/3y)-2/3y`
=`2-4/3y`
so
`y=-3/4(x^2-2)`