JEE Main 2022 (Online) 27th July Evening ShiftQuestion 1
For the curve $C:(x^2+y^2-3)+(x^2-y^2-1)^5=0$, the value of $3y'-y^3y''$, at the point $(\alpha,\alpha)$, $\alpha>0$, on $C$, is equal to ________.
Differentiation and its Application
Answer$16$
Solution
Concept. Find $\alpha$, differentiate the curve twice implicitly, and put the point in at each stage. Write $w=x^2-y^2-1$. Formulas used. (1) $\dfrac{d}{dx}w^5=5w^4\,w'$ with $w'=2x-2yy'$Step 1 - The point. At $(\alpha,\alpha)$, $w=-1$: $2\alpha^2-3-1=0$, so $\alpha=\sqrt2$. Step 2 - First derivative. By (1), and dividing by $2$: $$x\left(1+5w^4\right)+yy'\left(1-5w^4\right)=0.$$ At the point $w^4=1$: $6\sqrt2-4\sqrt2\,y'=0$, so $y'=\frac32$. Step 3 - Differentiate again. $$\left(1+5w^4\right)+20xw^3w'+\left(y'^2+yy''\right)\left(1-5w^4\right)-20yy'w^3w'=0.$$ Step 4 - Put in values. $w^3=-1$, $w'=2\sqrt2-2\sqrt2\cdot\frac32=-\sqrt2$: $$6+40-4\left(\tfrac94+\sqrt2\,y''\right)-60=0\ \Longrightarrow\ y''=-\frac{23}{4\sqrt2}.$$ Step 5 - Value. $y^3=2\sqrt2$, so $$3y'-y^3y''=\frac92+2\sqrt2\cdot\frac{23}{4\sqrt2}=\frac92+\frac{23}{2}=16.$$ Answer: 16