Question

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

Answer 1

Answer: 16

$(\alpha ,\alpha )$ 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 $C$
$2x+2y\cdot y^{\prime }+5{\left(x^2-y^2-1\right)}^4\left(2x-2yy{y}^{\prime }\right)=\mathrm{0......}$(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,
$\left(1+\frac{9}{4}\right)+\sqrt{2}{y}^{\prime \prime }-40{\left(\sqrt{2}-\sqrt{2}\cdot \frac{3}{2}\right)}^2$
$+5(1)\left(1-\frac{9}{4}-\sqrt{2}{y}^{\prime \prime }\right)=0$
$\Rightarrow 4\sqrt{2}{y}^{\prime \prime }=-23$
$\therefore 3y^{\prime }-y^3{y}^{\prime \prime }=\frac{9}{2}+\frac{23}{2}=16$