Question

For the curve C: (x² + y² – 3) + (x² – y² – 1)⁵ = 0, the value of 3y' – y³ y", at the point (α, α), α < 0, on C, is equal to ______.

Options

• 15

• 16

• 17

• 18

Answer 1

For the curve C : (x² + y² – 3) + (x² – y² – 1)⁵ = 0, the value of 3y' – y³y", at the point (α, α), α < 0, on C, is equal to 16.

Explanation:

Given a curve C : x² + y² – 3 + (x² – y² – 1)⁵ = 0

Which satisfies the point (α, α).

2α² – 3 + –1⁵ = 0

`\implies` α = `sqrt(2)`

Now, differentiate w.r.t. x.

2x + 2y . y' + 5(x² – y² – 1)⁴ (2x – 2yy') = 0   ...(i)

So, point is `(sqrt(2), sqrt(2))`

`sqrt(2) + sqrt(2)y^'5(-1)^4 (sqrt(2) - sqrt(2)y^')` = 0

`\implies` y' = `3/2`  ...(ii)

Diff. (i) w.r.t. x.

Again, Diff. (i) w.r.t. x

1 + (y')² + yy'' + 20(x² – y² – 1)³ (x – yy')².2 + 5(x² – y² – 1)⁴ (1 – (y')² – yy'') = 0

At `(sqrt(2), sqrt(2))` and y' = `3/2`

Now, Take `(1 + 9/4) + sqrt(2)y^('') - 40(sqrt(2) - sqrt(2). 3/2)^2 + 5(1)(1 - 9/4 - sqrt(2)y^(''))` = 0

`\implies 4sqrt(2)y^('')` = –23

Therefore, 3y' – y3y" = `9/2 + 23/2` = 16