JEE Main 2022 — Differentiation Question with Solution

Given, 

$\left(\alpha ,\alpha \right)$ lies on $C:\left(x^2+y^2-3\right)+{\left(x^2-y^2-1\right)}^5=0$

So on putting $\left(\alpha ,\alpha \right)$ we get,

$2{\alpha }^2-3-1^5=0$

$\Rightarrow \alpha =\sqrt{2}$

Now, differentiating the curve $C$ we get,

$2x+2y·y^{\text{'}}+5{\left(x^2-y^2-1\right)}^4\left(2x-2y{y}^{\text{'}}\right)=0 \cdots \left(1\right)$

At $\left(\sqrt{2},\sqrt{2}\right)$

$\sqrt{2}+\sqrt{2}{y}^{\text{'}}+5{\left(-1\right)}^4\left(\sqrt{2}-\sqrt{2}{y}^{\text{'}}\right)=0$

$\Rightarrow y^{\text{'}}=\frac{3}{2} \cdots \left(2\right)$

Again, Diff. $\left(1\right)$ w.r.t. $x$ we get,

$1+{\left(y^{\text{'}}\right)}^2+yy{y}^{\text{'}\text{'}}+20{\left(x^2-y^2-1\right)}^3{\left(x-y{y}^{\text{'}}\right)}^2·2$

$+5{\left(x^2-y^2-1\right)}^4\left(1-{\left(y^{\text{'}}\right)}^2-yy{y}^{\text{'}\text{'}}\right)=0$

At $\left(\sqrt{2},\sqrt{2}\right)$ and $y^{\text{'}}=\frac{3}{2}$ 

We have,

$\left(1+\frac{9}{4}\right)+\sqrt{2}{y}^{\text{'}\text{'}}-40{\left(\sqrt{2}-\sqrt{2}·\frac{3}{2}\right)}^2$

$+5\left(1\right)\left(1-\frac{9}{4}-\sqrt{2}{y}^{\text{'}\text{'}}\right)=0$

$\Rightarrow 4\sqrt{2}{y}^{\text{'}\text{'}}=-23$

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

So, value of $3y^{\text{'}}-y^3{y}^{\text{'}\text{'}}=\frac{9}{2}+\frac{23}{2}=16$