Applications of Derivatives
Tangent and Differential Equations of Curves
nta_pyq_2023_apr
Grade 12
Question:
Let a curve $y=f(x),\ x\in(0,\infty)$ pass through the points $P\!\left(1,\dfrac{3}{2}\right)$ and $Q\!\left(a,\dfrac{1}{2}\right)$. If the tangent at any point $R(b,f(b))$ to the given curve cuts the $y$-axis at the point $S(0,c)$ such that $bc=3$, then $(PQ)^2$ is equal to _____.
Step-by-Step Solution
Key Concept: The condition $bc=3$ on the tangent-$y$-axis intercept translates to an ODE: $y - x\dfrac{dy}{dx} = \dfrac{3}{x}$, i.e., $d\!\left(\dfrac{y}{x}\right) = 3\,d(x^{-2}/(-2))$ after rearranging.
Tangent at $R$: $c=y-mx$ where $m=dy/dx$, and $bc=3$ gives $y-x\frac{dy}{dx}=\frac{3}{x}$. Rewriting: $d\!\left(\frac{y}{x}\right)=\frac{3}{-2}d(x^{-2})$, integrate to get $\frac{y}{x}=\frac{3}{2x^2}+C$. Using $P(1,\frac{3}{2})$: $C=0$, so $2xy=3$. Since $Q(a,\frac{1}{2})$ lies on curve: $a=3$. $(PQ)^2=(3-1)^2+(\frac{1}{2}-\frac{3}{2})^2=4+1=5$.
Correct Answer: 5