JEE Main 2021 (Online) 25th February Evening ShiftQuestion 1
If the curves $x=y^{4}$ and $xy=k$ cut at right angles, then $(4k)^{6}$ is equal to $\_\_\_\_\_\_$ .
Differentiation and its Application
Answer$4$
Solution
Concept. At the intersection the product of the two slopes is $-1$. Formulas used. (1) $x=y^4$: $\dfrac{dy}{dx}=\dfrac{1}{4y^3}$; $\ xy=k$: $\dfrac{dy}{dx}=-\dfrac yx$Step 1 - Intersection. $y^4\cdot y=k$, so $k=y^5$ and $x=y^4$. Then $-\frac yx=-\frac{1}{y^3}$. Step 2 - Right angle. By (1), $$\frac{1}{4y^3}\cdot\left(-\frac{1}{y^3}\right)=-1\ \Longrightarrow\ y^6=\frac14.$$Step 3 - Value. $k^6=y^{30}=\left(\frac14\right)^5=\frac{1}{1024}$, so $(4k)^6=\frac{4096}{1024}=4$. Answer: 4