For the curve C : (x^{2}+y^{2}-3)+(x^{2}-y^{2 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given the curve $C: (x^{2}+y^{2}-3)+(x^{2}-y^{2}-1)^{5}=0$.Since $(\alpha, \alpha)$ lies on $C$,we substitute $x=\alpha$ and $y=\alpha$:$(\alpha^{

Vedclass · Accepted Answer

$16$

Vedclass · Answer

(C) Given the curve $C: (x^{2}+y^{2}-3)+(x^{2}-y^{2}-1)^{5}=0$.Since $(\alpha, \alpha)$ lies on $C$,we substitute $x=\alpha$ and $y=\alpha$:$(\alpha^{2}+\alpha^{2}-3)+(\alpha^{2}-\alpha^{2}-1)^{5}=0$$(2\alpha^{2}-3)+(-1)^{5}=0$$2\alpha^{2}-3-1=0 \Rightarrow 2\alpha^{2}=4 \Rightarrow \alpha^{2}=2$. Since $\alpha > 0$,$\alpha = \sqrt{2}$.Differentiating the equation of the curve with respect to $x$:$2x + 2yy^{\prime} + 5(x^{2}-y^{2}-1)^{4}(2x - 2yy^{\prime}) = 0$.At $(\sqrt{2}, \sqrt{2})$:$2\sqrt{2} + 2\sqrt{2}y^{\prime} + 5(-1)^{4}(2\sqrt{2} - 2\sqrt{2}y^{\prime}) = 0$$2\sqrt{2} + 2\sqrt{2}y^{\prime} + 10\sqrt{2} - 10\sqrt{2}y^{\prime} = 0$$12\sqrt{2} - 8\sqrt{2}y^{\prime} = 0 \Rightarrow y^{\prime} = \frac{12}{8} = \frac{3}{2}$.Differentiating again:$2 + 2(y^{\prime})^{2} + 2yy^{\prime\prime} + 5[4(x^{2}-y^{2}-1)^{3}(2x-2yy^{\prime})^{2} + (x^{2}-y^{2}-1)^{4}(2-2(y^{\prime})^{2}-2yy^{\prime\prime})] = 0$.At $(\sqrt{2}, \sqrt{2})$ and $y^{\prime} = \frac{3}{2}$:$2 + 2(\frac{9}{4}) + 2\sqrt{2}y^{\prime\prime} + 5[4(-1)^{3}(2\sqrt{2}-2\sqrt{2}(\frac{3}{2}))^{2} + (-1)^{4}(2-2(\frac{9}{4})-2\sqrt{2}y^{\prime\prime})] = 0$$2 + \frac{9}{2} + 2\sqrt{2}y^{\prime\prime} + 5[-4(-\sqrt{2})^{2} + (2 - \frac{9}{2} - 2\sqrt{2}y^{\prime\prime})] = 0$$\frac{13}{2} + 2\sqrt{2}y^{\prime\prime} + 5[-8 - \frac{5}{2} - 2\sqrt{2}y^{\prime\prime}] = 0$$\frac{13}{2} + 2\sqrt{2}y^{\prime\prime} - 40 - \frac{25}{2} - 10\sqrt{2}y^{\prime\prime} = 0$$-8\sqrt{2}y^{\prime\prime} = 40 + \frac{25-13}{2} = 40 + 6 = 46 \Rightarrow y^{\prime\prime} = -\frac{46}{8\sqrt{2}} = -\frac{23}{4\sqrt{2}}$.Now,$3y^{\prime} - y^{3}y^{\prime\prime} = 3(\frac{3}{2}) - (\sqrt{2})^{3}(-\frac{23}{4\sqrt{2}}) = \frac{9}{2} - (2\sqrt{2})(-\frac{23}{4\sqrt{2}}) = \frac{9}{2} + \frac{23}{2} = \frac{32}{2} = 16$.

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

Similar Questions

For $x>1$,if $(2x)^{2y} = 4e^{2x-2y}$,then $(1+\log 2x)^2 \frac{dy}{dx}$ is equal to

If $x+y=\tan ^{-1} y$ and $\frac{d^{2} y}{d x^{2}}=f(y) \frac{d y}{d x}$,then $f(y)$ is equal to

If $3 f(\cos x) + 2 f(\sin x) = 5 x$,then $f^{\prime}(\cos x) + f^{\prime}(\sin x) =$

If $x^y \cdot y^x = 16$,then $\frac{dy}{dx}$ at $(2, 2)$ is

If $xy \neq 0, x+y \neq 0$ and $x^m y^n=(x+y)^{m+n}$,where $m, n \notin N$,then $\frac{dy}{dx}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module