JEE Main 2022 — Differentiation Question with Solution

$$\because$$ $$C:(x^2+y^2-3)+(x^2-y^2-1{)}^5=0$$ for point ($$\alpha$$, $$\alpha$$)

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

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

On differentiating $$(x^2+y^2-3)+(x^2-y^2-1{)}^5=0$$ we get

$$x+y{y}^{\prime }+5(x^2-y^2-1{)}^4(x-y{y}^{\prime })=0$$ ...... (i)

When $$x=y=\sqrt{2}$$ then $$y^{\prime }=\frac{3}{2}$$

Again on differentiating eq. (i) we get :

$$1+(y^{\prime }{)}^2+y{y}^{\prime \prime }+20(x^2-y^2-1)(2x-2y{y}^{\prime })(x-y^{\prime }y)+5(x^2-y^2-1{)}^4(1-y{}^{\prime 2}-y{y}^{\prime \prime })=0$$

For $$x=y=\sqrt{2}$$ and $$y^{\prime }=\frac{3}{2}$$ we get $$y^{\prime \prime }=-\frac{23}{4\sqrt{2}}$$

$$\therefore$$ $$3y^{\prime }-y^3{y}^{\prime \prime }=3.\frac{3}{2}-{\left(\sqrt{2}\right)}^3.\left(-\frac{23}{4\sqrt{2}}\right)=16$$