If the curves x=y^{4} and xy=k cut at right a | vedclass

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(C) Given curves are $x=y^{4}$ and $xy=k$. For the intersection point,substitute $x=y^{4}$ into $xy=k$: $y^{4} \cdot y = k \Rightarrow y^{5} = k \ldot

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(C) Given curves are $x=y^{4}$ and $xy=k$. For the intersection point,substitute $x=y^{4}$ into $xy=k$: $y^{4} \cdot y = k \Rightarrow y^{5} = k \ldots(1)$. Now,differentiate $x=y^{4}$ with respect to $x$: $1 = 4y^{3} \frac{dy}{dx} \Rightarrow \frac{dy}{dx} = \frac{1}{4y^{3}}$. Next,differentiate $xy=k$ with respect to $x$: $y + x \frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = -\frac{y}{x}$. Since $x = \frac{k}{y}$,we have $\frac{dy}{dx} = -\frac{y}{k/y} = -\frac{y^{2}}{k}$. Since the curves intersect at right angles,the product of their slopes is $-1$: $\left(\frac{1}{4y^{3}}\right) \cdot \left(-\frac{y^{2}}{k}\right) = -1$. $\Rightarrow \frac{1}{4yk} = 1 \Rightarrow y = \frac{1}{4k}$. Substitute $y = \frac{1}{4k}$ into equation $(1)$: $\left(\frac{1}{4k}\right)^{5} = k$. $\Rightarrow \frac{1}{(4k)^{5}} = k$. $\Rightarrow (4k)^{6} = 4$.

If the curves $x=y^{4}$ and $xy=k$ cut at right angles,then $(4k)^{6}$ is equal to ..... .

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