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

x = y^4 and xy = k for intersection y^5 = k ..... (1) Also x = y^4 => 1 = 4y^3dy / dx => dy / dx = 1 / 4y^3 for xy = k => x = k / y => 1 = - k / y^2.dy / dx => dy / dx = - y^2 / k Because Curve cut orthogonally => 1 / 4y^3 x ( - y^2 / k ) = - 1 => y = 1 / 4k Therefore from (1), y^5 = k => 1 / (4k)^5 = k => 4 = (4k)^6

If the curves x y4 and xy k cut at right angles then 4k... | null (Mathematics)

x = {y^4} \text{ and } xy = k

\text{for intersection} \quad {y^5} = k \quad ..... (1)

\text{ Also } \quad x = {y^4}

\Rightarrow 1 = 4{y^3}{{dy} \over {dx}} \Rightarrow {{dy} \over {dx}} = {1 \over {4{y^3}}}

\text{ for } \quad xy = k \Rightarrow x = {k \over y}

\Rightarrow 1 = - {k \over {{y^2}}}.{{dy} \over {dx}}

\Rightarrow {{dy} \over {dx}} = {{ - {y^2}} \over k}

\because

Curve cut orthogonally

\Rightarrow {1 \over {4{y^3}}} \times \left( {{{ - {y^2}} \over k}} \right) = - 1

\Rightarrow y = {1 \over {4k}}

\therefore \text{from (1),} \quad {y^5} = k

\Rightarrow {1 \over {{{(4k)}^5}}} = k

\Rightarrow 4 = {(4k)^6}

Explanation

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