Question

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

Answer 1

$(\alpha, \alpha)$ lies on
$ C:\left(x^2+y^2-3\right)+\left(x^2-y^2-1\right)^5=0$
Put $(\alpha, \alpha),\left(2 \alpha^2-3\right)+(-1)^5=0 $
$\Rightarrow \alpha=\sqrt{2}$
Now, differentiate $C$
$ 2 x +2 y \cdot y ^{\prime}+5\left( x ^2- y ^2-1\right)^4\left(2 x -2 yy y ^{\prime}\right)=0 ......$(1)
At $(\sqrt{2}, \sqrt{2})$
$ \sqrt{2}+\sqrt{2} y ^{\prime}+5(-1)^4\left(\sqrt{2}-\sqrt{2} y ^{\prime}\right)=0 $
$ \Rightarrow y ^{\prime}=\frac{3}{2}.....$(2)
Diff. (1) w.r.t. $x$
Again, Diff. (1) w.r.t. x
$1+\left(y^{\prime}\right)^2+y y^{\prime \prime}+20\left(x^2-y^2-1\right)^3\left(x-y y^{\prime}\right)^2 \cdot 2$$+5\left(x^2-y^2-1\right)^4\left(1-\left(y^{\prime}\right)^2-y y^{\prime \prime}\right)=0$
At $(\sqrt{2}, \sqrt{2})$ and $y ^{\prime}=\frac{3}{2}$
We have,
$ \left(1+\frac{9}{4}\right)+\sqrt{2} y^{\prime \prime}-40\left(\sqrt{2}-\sqrt{2} \cdot \frac{3}{2}\right)^2 $
$ +5(1)\left(1-\frac{9}{4}-\sqrt{2} y^{\prime \prime}\right)=0 $
$ \Rightarrow 4 \sqrt{2} y^{\prime \prime}=-23 $
$ \therefore 3 y^{\prime}-y^3 y^{\prime \prime}=\frac{9}{2}+\frac{23}{2}=16$