JEE Main 2024 (Online) 30th January Evening ShiftQuestion 1
Let $Y=Y(X)$ be a curve lying in the first quadrant such that the area enclosed by the line $Y-y=Y'(x)(X-x)$ and the co-ordinate axes, where $(x,y)$ is any point on the curve, is always $\frac{-y^{2}}{2Y'(x)}+1$, $Y'(x)\neq0$. If $Y(1)=1$, then $12Y(2)$ equals $\_\_\_\_\_\_$ .
Differential Equation
Answer$20$
Solution
Concept. Write the tangent's intercepts, set the area equal to the given expression, and solve for $y'$. Formulas used. (1) Tangent at $(x,y)$: $X$-intercept $x-\frac{y}{y'}$, $Y$-intercept $y-xy'$(2) $\left(\dfrac{y}{x^2}\right)'=\dfrac{x^2y'-2xy}{x^4}$Step 1 - Area. By (1), $x-\frac y{y'}=-\frac{y-xy'}{y'}$, so the area $\frac12(\text{product of intercepts})$ is $$-\frac{(y-xy')^2}{2y'}.$$ Step 2 - Equate. $-\dfrac{(y-xy')^2}{2y'}=-\dfrac{y^2}{2y'}+1$. Multiply by $-2y'$: $$(y-xy')^2=y^2-2y'\ \Longrightarrow\ y'\left(x^2y'-2xy+2\right)=0.$$ Since $y'\ne0$: $x^2y'-2xy=-2$. Step 3 - Solve. By (2), $\left(\frac{y}{x^2}\right)'=-\frac{2}{x^4}$, so $\frac{y}{x^2}=\frac{2}{3x^3}+C$, i.e. $y=\frac{2}{3x}+Cx^2$. $Y(1)=1$ gives $C=\frac13$. Step 4 - Value. $Y(2)=\frac13+\frac43=\frac53$, so $12Y(2)=20$. Answer: 20