0
You visited us 0 times! Enjoying our articles? Unlock Full Access!
Question

Let $$x^k+y^k=a^k,(a,k > 0)$$ and
$$\dfrac{dy}{dx}+\left(\dfrac{y}{x}\right)^{1/3}=0$$, then $$k$$ is :

A
$$\dfrac{4}{3}$$
B
$$\dfrac{1}{3}$$
C
$$\dfrac{2}{3}$$
D
$$\dfrac{3}{2}$$
Solution
Verified by Toppr

Correct option is B. $$\dfrac{2}{3}$$
$$x^k+y^k=a^k$$
differentiating w.r.t. x
$$kx^{k-1}+k.y^{k-1}\dfrac{dy}{dx}=0$$
$$\dfrac{dy}{dx}=-\dfrac{kx^{k-1}}{k.y^{k-1}}$$
$$\dfrac{dy}{dx}=-\left(\dfrac{x}{y}\right)^{k-1}$$
$$\dfrac{dy}{dx}+\left(\dfrac{x}{y}\right)^{k-1}=0$$
$$k-1=\dfrac{-1}{3}$$
$$\boxed{k=\dfrac{2}{3}}$$

Was this answer helpful?
11
Similar Questions
Q1
Let $$x^k+y^k=a^k,(a,k > 0)$$ and
$$\dfrac{dy}{dx}+\left(\dfrac{y}{x}\right)^{1/3}=0$$, then $$k$$ is :
View Solution
Q2
Let xk+yk=ak,(a,k>0) and dydx+(yx)13=0, then k is
View Solution
Q3
Let xk+yk=ak,(a,k>0) and dydx+(yx)13=0, then k is
View Solution
Q4

Let xk+yk=ak, a,k>0 and dydx+yx13=0, then k is


View Solution
Q5
Let a=ik,b=xi+j+(1x)k and c=yi+xj+(1+xy)k. Then, [abc] depends on
View Solution