For the curve $C: (x^2 + y^2 - 3) + (x^2 - y^2 - 1)^5 = 0$, the value of $3y' - y^3 y''$, at the point $(\alpha, \alpha)$, $\alpha > 0$, on $C$, is equal to ____.

JEE Main · 2022 · 27 Jul 2022 · Shift 2 · Continuity and Differentiability

For the curve

C : (x^{2} + y^{2} - 3) + (x^{2} - y^{2} - 1)^{5} = 0

, the value of

3 y^{'} - y^{3} y^{''}

, at the point

(α, α)

α > 0

, on

C

, is equal to ____.

Answer: 16

Substituting

(α, α)

into the curve equation:

2 α^{2} - 3 + (- 1)^{5} = 0 \Rightarrow α = 2

.

Differentiating the curve equation:

2 x + 2 y y^{'} + 5 (x^{2} - y^{2} - 1)^{4} (2 x - 2 y y^{'}) = 0

.

At

(2, 2)

2 + 2 y^{'} + 5 (- 1)^{4} (2 - 2 y^{'}) = 0

, giving

y^{'} = \frac{3}{2}

.

Differentiating again:

1 + (y^{'})^{2} + y y^{''} + 20 (x^{2} - y^{2} - 1)^{3} (x - y y^{'})^{2} + 5 (x^{2} - y^{2} - 1)^{4} (1 - (y^{'})^{2} - y y^{''}) = 0

.

At

(2, 2)

with

y^{'} = \frac{3}{2}

, solving gives

y^{''} = - \frac{23}{4 2}

.

Therefore:

3 y^{'} - y^{3} y^{''} = \frac{9}{2} - 22 \cdot (- \frac{23}{4 2}) = \frac{9}{2} + \frac{23}{2} = 16

JEE Main · 2022 · 27 Jul 2022 · Shift 2 · Integer answer